鑷充簬bfs鐨勫仛娉曞涓嬶細
1錛屽紑濮嬬偣榪涙爤
2錛屽紑濮嬬偣鑳界洿鎺ュ埌杈?涓嶈姳鏃墮棿鐨?鐨勬墍鏈夌殑鐐硅繘鏍?br /> 3錛屽垎涓ゆ
3.1 寮濮嬬偣涓嶈兘鐩存帴鍒拌揪錛堣鏃墮棿錛夌殑鐐硅繘鏍?br /> 3.2 灝嗚繖涓偣鑳界洿鎺ュ埌杈?涓嶈姳鏃墮棿鐨?鐨勬墍鏈夌殑鐐硅繘鏍?br /> 3.3 璺寵漿鍒?
4 璺寵漿鍒?
娉細寮濮嬬偣涓哄綋鍓嶅嚭鏍堢殑絎竴涓偣
涓嶆槑鐧借繖涓繃紼嬬殑鍙互鐪嬩唬鐮侊紙铏界劧浠ｇ爜鏈夌偣……錛岃繕鍙互榪涗竴姝ョ畝鍖栵紝 浠ュ悗涓嶈兘鑰佹兂鐫浼樺厛浜嗭紝璋佷紭鍏堣皝鏉叿錛屽綋鐒朵笉鏄鎴戜滑鐨勫箍澶уコ鍚岃優……錛?br />
1 #include <iostream>
2 #include <queue>
3 using namespace std;
4 struct node
5 {
6 int x, y, time;
7 /*friend bool operator < (node a, node b)
8 {
9 return a.time > b.time;
10 }*/
11 };
12 int n, m;
13 int xi[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
14 int yi[8] = {0, 1, 1, 1, 0, -1, -1, -1};
15 char map[1005][1005];
16 bool vist[1005][1005];
17 bool check(int dx, int dy)
18 {
19 if (dx >= 0 && dy >=0 && dx < n && dy < m) return true;
20 return false;
21 }
22 queue <node> q;
23 int bfs(node now, node end)
24 {
25 while (!q.empty())q.pop();
26 now.time = 0;
27 q.push(now);
28
29 for (int i = 0; i < n; i++)
30 for (int j = 0; j < m; j++)
31 vist[i][j] = false;
32 vist[now.x][now.y] = true;
33 node next, tmp;
34 bool flag = false;
35 while (!q.empty())
36 {
37 now = q.front();
38 tmp = now;
39 if (now.x == end.x && now.y == end.y) return now.time;
40 q.pop();
41 flag = false;
42 while (1)
43 {
44 next.x = tmp.x + xi[map[tmp.x][tmp.y]-'0'];
45 next.y = tmp.y + yi[map[tmp.x][tmp.y]-'0'];
46 if (check(next.x, next.y) && !vist[next.x][next.y])
47 {
48 if (next.x == end.x && next.y == end.y) return tmp.time;
49 next.time = tmp.time;
50 vist[next.x][next.y] = true;
51 q.push(next);
52 tmp = next;
53 }else break;
54 }
55 for (int i = 0; i < 8; i++)
56 {
57 next.x = now.x + xi[i];
58 next.y = now.y + yi[i];
59 if (check(next.x, next.y) && !vist[next.x][next.y])
60 {
61 if (map[now.x][now.y] - '0' == i) next.time = now.time;
62 else next.time = now.time + 1;
63 if (next.x == end.x && next.y == end.y) return next.time;
64 vist[next.x][next.y] = true;
65 q.push(next);
66 tmp = next;
67 while (1)
68 {
69 next.x = tmp.x + xi[map[tmp.x][tmp.y]-'0'];
70 next.y = tmp.y + yi[map[tmp.x][tmp.y]-'0'];
71 if (check(next.x, next.y) && !vist[next.x][next.y])
72 {
73 if (next.x == end.x && next.y == end.y) return tmp.time;
74 next.time = tmp.time;
75 vist[next.x][next.y] = true;
76 q.push(next);
77 tmp = next;
78 }else break;
79 }
80 }
81 }
82 }
83 return 0;
84 }
85 int main()
86 {
87 while (scanf("%d%d", &n, &m) != EOF)
88 {
89 int i, j;
90 for (i = 0; i < n; i++)
91 scanf("%s", map[i]);
92 int T;
93 scanf("%d", &T);
94 node now, end;
95 for (i = 0; i < T; i++)
96 {
97 scanf("%d%d%d%d", &now.x, &now.y, &end.x, &end.y);
98 now.time = 0;
99 now.x --;
100 now.y --;
101 end.x --;
102 end.y --;
103 if (now.x == end.x && now.y == end.y)
104 {
105 printf("0\n");
106 }else printf("%d\n", bfs(now, end));
107 }
108 }
109 return 0;
110 }
111
2 #include <queue>
3 using namespace std;
4 struct node
5 {
6 int x, y, time;
7 /*friend bool operator < (node a, node b)
8 {
9 return a.time > b.time;
10 }*/
11 };
12 int n, m;
13 int xi[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
14 int yi[8] = {0, 1, 1, 1, 0, -1, -1, -1};
15 char map[1005][1005];
16 bool vist[1005][1005];
17 bool check(int dx, int dy)
18 {
19 if (dx >= 0 && dy >=0 && dx < n && dy < m) return true;
20 return false;
21 }
22 queue <node> q;
23 int bfs(node now, node end)
24 {
25 while (!q.empty())q.pop();
26 now.time = 0;
27 q.push(now);
28
29 for (int i = 0; i < n; i++)
30 for (int j = 0; j < m; j++)
31 vist[i][j] = false;
32 vist[now.x][now.y] = true;
33 node next, tmp;
34 bool flag = false;
35 while (!q.empty())
36 {
37 now = q.front();
38 tmp = now;
39 if (now.x == end.x && now.y == end.y) return now.time;
40 q.pop();
41 flag = false;
42 while (1)
43 {
44 next.x = tmp.x + xi[map[tmp.x][tmp.y]-'0'];
45 next.y = tmp.y + yi[map[tmp.x][tmp.y]-'0'];
46 if (check(next.x, next.y) && !vist[next.x][next.y])
47 {
48 if (next.x == end.x && next.y == end.y) return tmp.time;
49 next.time = tmp.time;
50 vist[next.x][next.y] = true;
51 q.push(next);
52 tmp = next;
53 }else break;
54 }
55 for (int i = 0; i < 8; i++)
56 {
57 next.x = now.x + xi[i];
58 next.y = now.y + yi[i];
59 if (check(next.x, next.y) && !vist[next.x][next.y])
60 {
61 if (map[now.x][now.y] - '0' == i) next.time = now.time;
62 else next.time = now.time + 1;
63 if (next.x == end.x && next.y == end.y) return next.time;
64 vist[next.x][next.y] = true;
65 q.push(next);
66 tmp = next;
67 while (1)
68 {
69 next.x = tmp.x + xi[map[tmp.x][tmp.y]-'0'];
70 next.y = tmp.y + yi[map[tmp.x][tmp.y]-'0'];
71 if (check(next.x, next.y) && !vist[next.x][next.y])
72 {
73 if (next.x == end.x && next.y == end.y) return tmp.time;
74 next.time = tmp.time;
75 vist[next.x][next.y] = true;
76 q.push(next);
77 tmp = next;
78 }else break;
79 }
80 }
81 }
82 }
83 return 0;
84 }
85 int main()
86 {
87 while (scanf("%d%d", &n, &m) != EOF)
88 {
89 int i, j;
90 for (i = 0; i < n; i++)
91 scanf("%s", map[i]);
92 int T;
93 scanf("%d", &T);
94 node now, end;
95 for (i = 0; i < T; i++)
96 {
97 scanf("%d%d%d%d", &now.x, &now.y, &end.x, &end.y);
98 now.time = 0;
99 now.x --;
100 now.y --;
101 end.x --;
102 end.y --;
103 if (now.x == end.x && now.y == end.y)
104 {
105 printf("0\n");
106 }else printf("%d\n", bfs(now, end));
107 }
108 }
109 return 0;
110 }
111
]]>
鍏蜂綋鐪嬩唬鐮侊紝涓嶅仛榪囧鐨勮В閲娿?
1 #include <iostream>
2 #include <queue>
3 using namespace std;
4 struct node
5 {
6 int x, y;
7 int step, dir;
8 };
9 int n, m;
10 int xi[4] = {0, 1, -1, 0};
11 int yi[4] = {1, 0, 0, -1};
12 int vist[82][82][4];
13 char map[82][82];
14 node start;
15 queue <node> q;
16 bool check(int dx, int dy)
17 {
18 if(dx >= 1 && dx <= n && dy >= 1 && dy <= m) return true;
19 else return false;
20 }
21 bool find(node a)
22 {
23 if ((a.x == 1 || a.x == n || a.y == 1 || a.y == m)) return true;
24 else return false;
25 }
26 int bfs()
27 {
28 while (!q.empty())q.pop();
29 memset(vist, 0, sizeof(vist));
30 start.dir = -1;
31 start.step = 0;
32 q.push(start);
33 node now, next;
34 bool flag = true;
35 int tmp;
36 while (!q.empty())
37 {
38 now = q.front();
39 q.pop();
40 if (find(now)) return now.step;
41
42 flag = false;
43 for (int i = 0 ; i < 4; i++)
44 {
45 if (now.dir == i) continue;
46 if (now.dir >=0 && 3-now.dir == i) continue;
47 next.x = now.x + xi[i];
48 next.y = now.y + yi[i];
49 next.step = now.step + 1;
50 if (check(next.x, next.y) && map[next.x][next.y] == '.' )
51 {
52 if (vist[next.x][next.y][i] == 0)
53 {
54 vist[next.x][next.y][i] = next.step;
55 next.dir = i;
56 q.push(next);
57 }
58 else if (vist[next.x][next.y][i] > next.step)
59 {
60 vist[next.x][next.y][i] = next.step;
61 next.dir = i;
62 q.push(next);
63 }
64 flag = true;
65 }
66 }
67 if (!flag)
68 {
69 int i = now.dir;
70 if (i < 0) return 0;
71 next.x = now.x + xi[i];
72 next.y = now.y + yi[i];
73 next.step = now.step + 1;
74 if (check(next.x, next.y) && map[next.x][next.y] == '.' )
75 {
76 if (vist[next.x][next.y][i] == 0)
77 {
78 vist[next.x][next.y][i] = next.step;
79 next.dir = i;
80 q.push(next);
81 }
82 else if (vist[next.x][next.y][i] > next.step)
83 {
84 vist[next.x][next.y][i] = next.step;
85 next.dir = i;
86 q.push(next);
87 }
88 flag = true;
89 }
90 }
91 }
92 return -1;
93 }
94 int main()
95 {
96 int cas;
97 scanf("%d", &cas);
98 while (cas--)
99 {
100 scanf("%d%d", &n, &m);
101 int i, j;
102 for (i = 1; i <= n; i++)
103 scanf("%s", map[i]+1);
104 for (i = 1; i <= n; i++)
105 for (j = 1; j <= m; j++)
106 {
107 if (map[i][j] == '@')
108 {
109 start.x = i;
110 start.y = j;
111 map[i][j] = '.';
112 }
113 }
114 int ans = bfs();
115 printf("%d\n", ans);
116 }
117 return 0;
118 }
119
2 #include <queue>
3 using namespace std;
4 struct node
5 {
6 int x, y;
7 int step, dir;
8 };
9 int n, m;
10 int xi[4] = {0, 1, -1, 0};
11 int yi[4] = {1, 0, 0, -1};
12 int vist[82][82][4];
13 char map[82][82];
14 node start;
15 queue <node> q;
16 bool check(int dx, int dy)
17 {
18 if(dx >= 1 && dx <= n && dy >= 1 && dy <= m) return true;
19 else return false;
20 }
21 bool find(node a)
22 {
23 if ((a.x == 1 || a.x == n || a.y == 1 || a.y == m)) return true;
24 else return false;
25 }
26 int bfs()
27 {
28 while (!q.empty())q.pop();
29 memset(vist, 0, sizeof(vist));
30 start.dir = -1;
31 start.step = 0;
32 q.push(start);
33 node now, next;
34 bool flag = true;
35 int tmp;
36 while (!q.empty())
37 {
38 now = q.front();
39 q.pop();
40 if (find(now)) return now.step;
41
42 flag = false;
43 for (int i = 0 ; i < 4; i++)
44 {
45 if (now.dir == i) continue;
46 if (now.dir >=0 && 3-now.dir == i) continue;
47 next.x = now.x + xi[i];
48 next.y = now.y + yi[i];
49 next.step = now.step + 1;
50 if (check(next.x, next.y) && map[next.x][next.y] == '.' )
51 {
52 if (vist[next.x][next.y][i] == 0)
53 {
54 vist[next.x][next.y][i] = next.step;
55 next.dir = i;
56 q.push(next);
57 }
58 else if (vist[next.x][next.y][i] > next.step)
59 {
60 vist[next.x][next.y][i] = next.step;
61 next.dir = i;
62 q.push(next);
63 }
64 flag = true;
65 }
66 }
67 if (!flag)
68 {
69 int i = now.dir;
70 if (i < 0) return 0;
71 next.x = now.x + xi[i];
72 next.y = now.y + yi[i];
73 next.step = now.step + 1;
74 if (check(next.x, next.y) && map[next.x][next.y] == '.' )
75 {
76 if (vist[next.x][next.y][i] == 0)
77 {
78 vist[next.x][next.y][i] = next.step;
79 next.dir = i;
80 q.push(next);
81 }
82 else if (vist[next.x][next.y][i] > next.step)
83 {
84 vist[next.x][next.y][i] = next.step;
85 next.dir = i;
86 q.push(next);
87 }
88 flag = true;
89 }
90 }
91 }
92 return -1;
93 }
94 int main()
95 {
96 int cas;
97 scanf("%d", &cas);
98 while (cas--)
99 {
100 scanf("%d%d", &n, &m);
101 int i, j;
102 for (i = 1; i <= n; i++)
103 scanf("%s", map[i]+1);
104 for (i = 1; i <= n; i++)
105 for (j = 1; j <= m; j++)
106 {
107 if (map[i][j] == '@')
108 {
109 start.x = i;
110 start.y = j;
111 map[i][j] = '.';
112 }
113 }
114 int ans = bfs();
115 printf("%d\n", ans);
116 }
117 return 0;
118 }
119
]]>
濡傛灉鑳藉鍒拌揪錛屽垯姹傚嚭瀵逛簬姣忎竴涓偣鍒板叾浠栨墍鏈夌殑鐐規渶鐭窛紱諱箣鍜岋紝灝嗚繖浜涘拰鐩稿姞灝辨槸鏈鍚庣殑緇撴灉
瑙ｉ鎬濊礬錛?br /> 瀵規瘡涓涓偣姹備竴嬈″崟婧愭渶鐭礬錛岀劧鍚庢眰鍜岋紝鐩稿姞鐨勫埌鏈鍚庣殑緇撴灉……
浣嗭紝綆椾竴涓嬫椂闂村鏉傚害錛?澶嶆潅搴︽槸O(M*N*M)銆傜敱浜嶮*N*M=3000*100*3000=9*108錛屼笉鍙兘AC
鎵浠ワ紝闇瑕佹垜浠彟杈熶粬寰勩傜綉涓婃湁璇村緩浠涔堟渶鐭礬寰勬爲錛岃繖涓垜涓嶆噦……
鎴戠殑鎬濊礬鏄細寮曠敤鍓嶉潰鐨勬濊礬錛屽姣忎竴涓偣姹備竴嬈℃渶鐭礬錛屽茍灝嗗叾姹傚拰錛屼繚瀛樺湪涓涓暟緇勯噷澶達紝瀹氫負sum[i]錛宨琛ㄧず鐫涓涓偣鍒版墍鏈夊叾浠栫偣鏈鐭礬涔嬪拰銆傚茍灝嗚繖浜涘拰鐩稿姞 ans = sum[1] + …… + sum[n];
鐒跺悗錛屽垹闄や竴鏉¤竟錛屽叾欏剁偣鏆傚畾涓簎,v錛屽榪欐潯杈圭殑涓涓《鐐箄鍦ㄤ竴嬈℃眰鏈鐭礬錛屽鏋滆繖涓偣錛屼笉鑳藉埌杈捐繖鏉¤竟鐨勫彟涓涓偣v錛屽垯 鐩存帴杈撳嚭INF
濡傛灉錛岃兘澶熷埌杈撅紝鍒欏v涔熸眰涓嬈℃渶鐭礬錛屽浜巙錛寁涓ょ偣鏉ヨ錛屾眰寰梪鍒版瘡涓涓偣鐨勬渶鐭礬涔嬪拰sum_u錛屾眰寰梫鍒版瘡涓涓偣鐨勬渶鐭礬涔嬪拰sum_v錛?br /> 鏈鍚庣粨鏋滀負錛?ans = ans + sum_u + sum_v - sum[u] - sum[v];
ans涓烘渶鍚庣瓟妗堬紙鍗冧竾璁頒綇鏄瘡涓緇勭殑錛屽洜涓烘湁m鏉¤竟錛?br /> 鍏蜂綋鐪嬩唬鐮侊細
http://acm.hdu.edu.cn/showproblem.php?pid=2433
鏃墮棿鏄?953ms
1 #include <iostream>
2 #include <queue>
3 using namespace std;
4 const int SIZE = 102;
5 const int INF = 1 << 30;
6 struct node
7 {
8 int v, w, next;
9 }map[SIZE * SIZE];
10
11 int head[SIZE], use;
12 int num[SIZE * 30];
13 int sum[SIZE];
14
15 void add(int u, int v, int w)
16 {
17 map[use].v = v;
18 map[use].w = w;
19 map[use].next = head[u];
20 head[u] = use ++;
21 }
22 void init()
23 {
24 use = 0;
25 memset(head, -1, sizeof(head));
26 memset(sum, 0, sizeof(sum));
27 }
28
29 queue <int> q;
30 bool vist[SIZE];
31 int dis[SIZE];
32
33 int spfa(int s)
34 {
35 while (!q.empty()) q.pop();
36
37 for (int i = 0; i < SIZE ; i ++) dis[i] = INF;
38
39 memset(vist, false, sizeof(vist));
40 dis[s] = 0;
41 vist[s] = true;
42 q.push(s);
43 int ans = 0;
44 while (!q.empty())
45 {
46 int u = q.front();
47 q.pop();
48 vist[u] = false;
49 ans += dis[u];
50 for (int i = head[u]; i != -1; i = map[i].next)
51 {
52 int v = map[i].v;
53 if (dis[v] > dis[u] + map[i].w)
54 {
55 dis[v] = dis[u] + map[i].w;
56 if (!vist[v])
57 {
58 vist[v] = true;
59 q.push(v);
60 }
61 }
62 }
63 }
64 return ans;
65 }
66 int main()
67 {
68 int n, m;
69 while (scanf("%d%d", &n, &m) != EOF)
70 {
71 int i, j, k;
72 int u, v;
73 init();
74 memset(num, 0, sizeof(num));
75 for (i = 1; i <= m; i++)
76 {
77 scanf("%d%d", &u, &v);
78 num[i] = use;
79 add(u, v, 1);
80 add(v, u, 1);
81 }
82 int ans = 0;
83 for (i = 1; i <= n; i++)
84 {
85 sum[i] = spfa(i);
86 ans += sum[i];
87 }
88 int tmp = ans;
89 for (i = 1; i <= m; i++)
90 {
91 map[num[i]].w = INF;
92 map[num[i]^1].w = INF;
93
94 ans = tmp;
95 ans += spfa(map[num[i]].v);
96
97 if (dis[map[num[i]^1].v] == INF)
98 {
99 printf("INF\n");
100 }
101 else
102 {
103 ans += spfa(map[num[i]^1].v);
104 ans = ans - sum[map[num[i]].v] - sum[map[num[i]^1].v];
105 printf("%d\n", ans);
106 }
107 map[num[i]].w = 1;
108 map[num[i]^1].w = 1;
109 }
110 }
111 return 0;
112 }
113
2 #include <queue>
3 using namespace std;
4 const int SIZE = 102;
5 const int INF = 1 << 30;
6 struct node
7 {
8 int v, w, next;
9 }map[SIZE * SIZE];
10
11 int head[SIZE], use;
12 int num[SIZE * 30];
13 int sum[SIZE];
14
15 void add(int u, int v, int w)
16 {
17 map[use].v = v;
18 map[use].w = w;
19 map[use].next = head[u];
20 head[u] = use ++;
21 }
22 void init()
23 {
24 use = 0;
25 memset(head, -1, sizeof(head));
26 memset(sum, 0, sizeof(sum));
27 }
28
29 queue <int> q;
30 bool vist[SIZE];
31 int dis[SIZE];
32
33 int spfa(int s)
34 {
35 while (!q.empty()) q.pop();
36
37 for (int i = 0; i < SIZE ; i ++) dis[i] = INF;
38
39 memset(vist, false, sizeof(vist));
40 dis[s] = 0;
41 vist[s] = true;
42 q.push(s);
43 int ans = 0;
44 while (!q.empty())
45 {
46 int u = q.front();
47 q.pop();
48 vist[u] = false;
49 ans += dis[u];
50 for (int i = head[u]; i != -1; i = map[i].next)
51 {
52 int v = map[i].v;
53 if (dis[v] > dis[u] + map[i].w)
54 {
55 dis[v] = dis[u] + map[i].w;
56 if (!vist[v])
57 {
58 vist[v] = true;
59 q.push(v);
60 }
61 }
62 }
63 }
64 return ans;
65 }
66 int main()
67 {
68 int n, m;
69 while (scanf("%d%d", &n, &m) != EOF)
70 {
71 int i, j, k;
72 int u, v;
73 init();
74 memset(num, 0, sizeof(num));
75 for (i = 1; i <= m; i++)
76 {
77 scanf("%d%d", &u, &v);
78 num[i] = use;
79 add(u, v, 1);
80 add(v, u, 1);
81 }
82 int ans = 0;
83 for (i = 1; i <= n; i++)
84 {
85 sum[i] = spfa(i);
86 ans += sum[i];
87 }
88 int tmp = ans;
89 for (i = 1; i <= m; i++)
90 {
91 map[num[i]].w = INF;
92 map[num[i]^1].w = INF;
93
94 ans = tmp;
95 ans += spfa(map[num[i]].v);
96
97 if (dis[map[num[i]^1].v] == INF)
98 {
99 printf("INF\n");
100 }
101 else
102 {
103 ans += spfa(map[num[i]^1].v);
104 ans = ans - sum[map[num[i]].v] - sum[map[num[i]^1].v];
105 printf("%d\n", ans);
106 }
107 map[num[i]].w = 1;
108 map[num[i]^1].w = 1;
109 }
110 }
111 return 0;
112 }
113
]]>
Description
On a grid map there are n little men and n houses. In each unit time, every little man can move one unit step, either horizontally, or vertically, to an adjacent point. For each little man, you need to pay a $1 travel fee for every step he moves, until he enters a house. The task is complicated with the restriction that each house can accommodate only one little man.
Your task is to compute the minimum amount of money you need to pay in order to send these n little men into those n different houses. The input is a map of the scenario, a '.' means an empty space, an 'H' represents a house on that point, and am 'm' indicates there is a little man on that point.
You can think of each point on the grid map as a quite large square, so it can hold n little men at the same time; also, it is okay if a little man steps on a grid with a house without entering that house.
Your task is to compute the minimum amount of money you need to pay in order to send these n little men into those n different houses. The input is a map of the scenario, a '.' means an empty space, an 'H' represents a house on that point, and am 'm' indicates there is a little man on that point.
You can think of each point on the grid map as a quite large square, so it can hold n little men at the same time; also, it is okay if a little man steps on a grid with a house without entering that house.
Input
There are one or more test cases in the input. Each case starts with a line giving two integers N and M, where N is the number of rows of the map, and M is the number of columns. The rest of the input will be N lines describing the map. You may assume both N and M are between 2 and 100, inclusive. There will be the same number of 'H's and 'm's on the map; and there will be at most 100 houses. Input will terminate with 0 0 for N and M.
Output
For each test case, output one line with the single integer, which is the minimum amount, in dollars, you need to pay.
Sample Input
2 2 .m H. 5 5 HH..m ..... ..... ..... mm..H 7 8 ...H.... ...H.... ...H.... mmmHmmmm ...H.... ...H.... ...H.... 0 0
Sample Output
2 10 28
KM綆楁硶錛?
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搴熻瘽涓嶅璇達紝鐪嬬▼搴忥細1 #include<iostream>
2 using namespace std;
3 #define Inf 10000000;
4 int map[105][105];
5 int slack[105];
6 int lx[105],ly[105];
7 bool x[105],y[105];
8 int link[105];
9 int n,m;
10 char mm[105][105];
11 bool dfs(int v)
12 {
13 x[v]=true;
14 int t;
15 for (int i=0;i<m;i++)
16 {
17 if (!y[i])
18 {
19 t=lx[v]+ly[i]-map[v][i];
20 if (t==0)
21 {
22 y[i]=true;
23 if (link[i]==-1||dfs(link[i]))
24 {
25 link[i]=v;
26 return true;
27 }
28 }
29 else slack[i]=min(slack[i],t);
30 }
31 }
32 return false;
33 }
34 void KM()
35 {
36 int i,j,k;
37 memset(link,-1,sizeof(link));
38 for (i=0;i<=m;i++)
39 {
40 lx[i]=-Inf;
41 ly[i]=0;
42 }
43 for (i=0;i<m;i++)
44 {
45 while (1)
46 {
47 memset(x,0,sizeof(x));
48 memset(y,0,sizeof(y));
49
50 for (j=0;j<m;j++) slack[j]=Inf;
51
52 if (dfs(i))break;
53
54 int d=Inf;
55 for (j=0;j<m;j++)
56 if (!y[j])d=min(slack[j],d);
57
58 for (j=0;j<m;j++)
59 {
60 if (x[j])lx[j]-=d;
61 if (y[j])ly[j]+=d;
62 }
63 for (j=0;j<m;j++)
64 if (!y[j])slack[j]-=d;
65 }
66 }
67 }
68 int main()
69 {
70 int i,j,k,t,l,v;
71 while (cin>>k>>t)
72 {
73 if (k+t==0)break;
74 for (i=1;i<=k;i++)
75 for (j=1;j<=t;j++)
76 scanf(" %c",&mm[i][j]);
77 memset(map,0,sizeof(map));
78 n=0,m=0;
79 for (i=1;i<=k;i++)
80 for (j=1;j<=t;j++)
81 {
82 if (mm[i][j]=='H')
83 {
84 n=0;
85 for (l=1;l<=k;l++)
86 for (v=1;v<=t;v++)
87 {
88 if (mm[l][v]=='m')
89 {
90 map[m][n]=-(abs(i-l)+abs(j-v));
91 n++;
92 }
93 }
94 m++;
95 }
96 }
97 KM();
98 int sum=0;
99 for (i=0;i<m;i++)
100 sum+=map[link[i]][i];
101 cout<<0-sum<<endl;
102 }
103 return 0;
104 }
105
106
]]>
]]>
#include<iostream>
using namespace std;
int n,m;//n涓簒闆嗗悎錛宮涓簓闆嗗悎
int map[305][305];//map[x][y]琛ㄧずx,y涓ょ偣涔嬮棿鏈夎竟鐩歌繛
bool visit[305];//璁板綍m涓殑i鑺傜偣鏄惁琚闂繃
int link[305];//璁板綍褰撳墠n涓殑i鑺傜偣鏄惁涓?nbsp;j鑺傜偣鐩歌繛
bool dfs(int a)
{
int i;
for (i=1;i<=n;i++)
{
if (map[a][i]&&!visit[i])
{
visit[i]=true;
if (link[i]==0||dfs(link[i]))
{
link[i]=a;
return true;
}
}
}
return false;
}
int find()
{
int sum=0;
for (int i=1;i<=n;i++)
{
memset(visit,0,sizeof(visit));
if (dfs(i))sum++;
}
return sum;//鏈澶у尮閰嶆暟
}
int main()
{
int t,i,j,k,x;
while (scanf("%d%d",&n,&m)!=EOF)
{
memset(map,0,sizeof(map));
memset(link,0,sizeof(link));
for (i=1;i<=n;i++)
{
scanf("%d",&j);
for (k=1;k<=j;k++)
{
scanf("%d",&x);
map[i][x]=1;
}
}
cout<<find()<<endl;
}
return 0;
}
using namespace std;
int n,m;//n涓簒闆嗗悎錛宮涓簓闆嗗悎
int map[305][305];//map[x][y]琛ㄧずx,y涓ょ偣涔嬮棿鏈夎竟鐩歌繛
bool visit[305];//璁板綍m涓殑i鑺傜偣鏄惁琚闂繃
int link[305];//璁板綍褰撳墠n涓殑i鑺傜偣鏄惁涓?nbsp;j鑺傜偣鐩歌繛
bool dfs(int a)
{
int i;
for (i=1;i<=n;i++)
{
if (map[a][i]&&!visit[i])
{
visit[i]=true;
if (link[i]==0||dfs(link[i]))
{
link[i]=a;
return true;
}
}
}
return false;
}
int find()
{
int sum=0;
for (int i=1;i<=n;i++)
{
memset(visit,0,sizeof(visit));
if (dfs(i))sum++;
}
return sum;//鏈澶у尮閰嶆暟
}
int main()
{
int t,i,j,k,x;
while (scanf("%d%d",&n,&m)!=EOF)
{
memset(map,0,sizeof(map));
memset(link,0,sizeof(link));
for (i=1;i<=n;i++)
{
scanf("%d",&j);
for (k=1;k<=j;k++)
{
scanf("%d",&x);
map[i][x]=1;
}
}
cout<<find()<<endl;
}
return 0;
}
]]>
1 #include<iostream>
2 using namespace std;
3 int map[12][12],tp[12][12],tt[12][12];
4 int n,m;
5 int Min=0xffffff,sum=0;
6 int x[4]={1,0,0,-1};
7 int y[4]={0,1,-1,0};
8 bool f=true;
9 //鏁扮粍鐨勪氦鎹?nbsp;
10 void fun(int a[12][12],int b[12][12])
11 {
12 for (int i=1;i<=n;i++)
13 for (int j=1;j<=n;j++)
14 a[i][j]=b[i][j];
15
16 }
17 void dfs(int x1,int y1,int sum,int p)
18 {
19 if(map[x1][y1]==3&&p>=0)
20 {
21 // 榪欓噷瑕佹敞鎰忥紝鎴戞槸浠?寮濮嬬殑錛屾悳鍒?鏃訛紝p搴旇鏄?浠ヤ笂錛?br>22 //鍒氬紑濮嬫槸娌℃悶娓呮錛宲澶т簬0,wa浜嗗嚑嬈★紝灝辨槸娌℃壘鍒伴敊璇紒
23 if(Min>sum)Min=sum;
24 //cout<<sum<<endl;
25 f=false;
26 return;
27 }
28 int dx,dy;
29 for (int i=0;i<4;i++)
30 {
31 dx=x1+x[i]; dy=y1+y[i];
32 if (map[dx][dy]!=0&&tp[dx][dy]==0&&p>=1)
33 {
34 if(map[dx][dy]==4)
35 {
36 map[dx][dy]=0;
37 int temp=p;
38 p=5;
39 // cout<<p<<' '<<dx<<' '<<dy<<endl;
40 //杈撳嚭璺緞錛屽亸浜庢煡鎵懼綋鍓嶇殑鍧愭爣浣嶇疆鍜屽墿浣欐椂闂磒
41 fun(tt,tp);
42 memset(tp,0,sizeof(tp));
43 //鍒?鏄彲浠ュ線鍥炴悳鐨勶紝鎵浠ュ墠闈㈢殑璧拌繃鐨勮礬寰勫簲璇ョЩ闄ゆ爣璁?br>44 //鐢ㄦ暟緇則t璁頒綇鍓嶉潰璧拌繃鐨勮礬寰勶紝浠ヤ究浜庡悗闈㈢殑鎼滅儲
45 tp[dx][dy]=1;
46 dfs(dx,dy,sum+1,p);
47 //鍑烘潵娣風殑錛屾槸瑕佽繕鐨勶紒榪欓噷涔熶竴鏍鳳紒
48 map[dx][dy]=4;
49 tp[dx][dy]=0;
50 p=temp;
51 fun(tp,tt);
52 }
53 else
54 {
55 tp[dx][dy]=1;
56 //cout<<"->"<<p<<' '<<dx<<' '<<dy<<endl;
57 //杈撳嚭璺緞錛屽亸浜庢煡鎵懼綋鍓嶇殑鍧愭爣浣嶇疆鍜屽墿浣欐椂闂磒
58 dfs(dx,dy,sum+1,p-1);
59 tp[dx][dy]=0;
60 }
61 }
62 }
63 }
64 int main()
65 {
66 int t;
67 cin>>t;
68 while (t--)
69 {
70 memset(map,0,sizeof(map));
71 memset(tp,0,sizeof(tp));
72 cin>>n>>m;
73 f=true;
74 int x1,y1,x2,y2;
75 for (int i=1;i<=n;i++)
76 for (int j=1;j<=m;j++)
77 {
78 cin>>map[i][j];
79 if(map[i][j]==2)x1=i,y1=j;
80 //if(map[i][j]==3)x2=i,y2=j;
81 }
82 Min=0xffffff,sum=0;
83 int p=5;
84 map[x1][y1]=0;
85 dfs(x1,y1,sum,5);
86 if(!f)cout<<Min<<endl;
87 else cout<<-1<<endl;
88 }
89 return 0;
90 }
91
2 using namespace std;
3 int map[12][12],tp[12][12],tt[12][12];
4 int n,m;
5 int Min=0xffffff,sum=0;
6 int x[4]={1,0,0,-1};
7 int y[4]={0,1,-1,0};
8 bool f=true;
9 //鏁扮粍鐨勪氦鎹?nbsp;
10 void fun(int a[12][12],int b[12][12])
11 {
12 for (int i=1;i<=n;i++)
13 for (int j=1;j<=n;j++)
14 a[i][j]=b[i][j];
15
16 }
17 void dfs(int x1,int y1,int sum,int p)
18 {
19 if(map[x1][y1]==3&&p>=0)
20 {
21 // 榪欓噷瑕佹敞鎰忥紝鎴戞槸浠?寮濮嬬殑錛屾悳鍒?鏃訛紝p搴旇鏄?浠ヤ笂錛?br>22 //鍒氬紑濮嬫槸娌℃悶娓呮錛宲澶т簬0,wa浜嗗嚑嬈★紝灝辨槸娌℃壘鍒伴敊璇紒
23 if(Min>sum)Min=sum;
24 //cout<<sum<<endl;
25 f=false;
26 return;
27 }
28 int dx,dy;
29 for (int i=0;i<4;i++)
30 {
31 dx=x1+x[i]; dy=y1+y[i];
32 if (map[dx][dy]!=0&&tp[dx][dy]==0&&p>=1)
33 {
34 if(map[dx][dy]==4)
35 {
36 map[dx][dy]=0;
37 int temp=p;
38 p=5;
39 // cout<<p<<' '<<dx<<' '<<dy<<endl;
40 //杈撳嚭璺緞錛屽亸浜庢煡鎵懼綋鍓嶇殑鍧愭爣浣嶇疆鍜屽墿浣欐椂闂磒
41 fun(tt,tp);
42 memset(tp,0,sizeof(tp));
43 //鍒?鏄彲浠ュ線鍥炴悳鐨勶紝鎵浠ュ墠闈㈢殑璧拌繃鐨勮礬寰勫簲璇ョЩ闄ゆ爣璁?br>44 //鐢ㄦ暟緇則t璁頒綇鍓嶉潰璧拌繃鐨勮礬寰勶紝浠ヤ究浜庡悗闈㈢殑鎼滅儲
45 tp[dx][dy]=1;
46 dfs(dx,dy,sum+1,p);
47 //鍑烘潵娣風殑錛屾槸瑕佽繕鐨勶紒榪欓噷涔熶竴鏍鳳紒
48 map[dx][dy]=4;
49 tp[dx][dy]=0;
50 p=temp;
51 fun(tp,tt);
52 }
53 else
54 {
55 tp[dx][dy]=1;
56 //cout<<"->"<<p<<' '<<dx<<' '<<dy<<endl;
57 //杈撳嚭璺緞錛屽亸浜庢煡鎵懼綋鍓嶇殑鍧愭爣浣嶇疆鍜屽墿浣欐椂闂磒
58 dfs(dx,dy,sum+1,p-1);
59 tp[dx][dy]=0;
60 }
61 }
62 }
63 }
64 int main()
65 {
66 int t;
67 cin>>t;
68 while (t--)
69 {
70 memset(map,0,sizeof(map));
71 memset(tp,0,sizeof(tp));
72 cin>>n>>m;
73 f=true;
74 int x1,y1,x2,y2;
75 for (int i=1;i<=n;i++)
76 for (int j=1;j<=m;j++)
77 {
78 cin>>map[i][j];
79 if(map[i][j]==2)x1=i,y1=j;
80 //if(map[i][j]==3)x2=i,y2=j;
81 }
82 Min=0xffffff,sum=0;
83 int p=5;
84 map[x1][y1]=0;
85 dfs(x1,y1,sum,5);
86 if(!f)cout<<Min<<endl;
87 else cout<<-1<<endl;
88 }
89 return 0;
90 }
91
]]>
銆銆榪欎釜鏂逛綅鐨勭‘瀹氫笌鐪嬪湴鍥劇殑鏂逛綅鏄竴鑷寸殑銆傜敱浜庤鎶?鈥?榪欎節涓暟濉湪閫傚綋鐨勬牸瀛愰噷錛岃繖涔濅釜鏁頒箣鍜屾槸45錛屾棤璁烘槸妯佺珫銆佹枩閮芥槸涓変釜鏁幫紝鎶?5騫沖潎鍒嗘垚涓夎錛屾瘡琛屼笁涓暟鐨勫拰閮芥槸15錛堝寘鎷í銆佺珫銆佹枩錛夈傛瘡涓変釜鏁扮殑鎯呭喌錛氭í鏈?縐嶏紝绔栨湁3縐嶏紝鏂滄湁2縐嶏紝鍏?縐嶃?/span>
鍙鐭ラ亾涓変釜鏁板氨鍙互鏋氫婦鎵鏈夌殑鏁頒簡錛?br>
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1 #include<iostream>
2 using namespace std;
3 int b[10],a[10];
4 int main(){
5 int f = 0;
6 for (int i=1;i<10;i++){
7 if(i!=5)b[1]=i;
8 for (int j=1;j<10;j++)
9 {
10 b[5] = 5;
11 if(j!=i&&j!=5&&i!=5){
12 b[2] = j;
13 b[8] = 15 - b[2] - b[5];
14 b[3] = 15 - b[1] - b[2];
15 b[9] = 15 - b[1] - b[5];
16 b[7] = 15 - b[3] - b[5];
17 b[4] = 15 - b[1] - b[7];
18 b[6] = 15 - b[3] - b[9];
19 if(b[4]+b[5]+b[6]==15&&b[7]+b[8]+b[9]==15)
20 {
21 f = 0;
22 memset(a,0,sizeof(a));
23 for (int k=1;k<10;k++) a[b[k]]++;
24 for (int k=1;k<10;k++) if(a[k]<=0||a[k]>1){f = 1;break;}
25 if(f==0){
26 for (int k=1;k<10;k++)
27 {
28 cout<< b[k] <<' ';
29 if(k%3==0)cout << endl;
30 }
31 cout << endl;
32 }
33 }
34 }
35 }
36 }
37 system("pause");
38 return 0;
39 }
40
2 using namespace std;
3 int b[10],a[10];
4 int main(){
5 int f = 0;
6 for (int i=1;i<10;i++){
7 if(i!=5)b[1]=i;
8 for (int j=1;j<10;j++)
9 {
10 b[5] = 5;
11 if(j!=i&&j!=5&&i!=5){
12 b[2] = j;
13 b[8] = 15 - b[2] - b[5];
14 b[3] = 15 - b[1] - b[2];
15 b[9] = 15 - b[1] - b[5];
16 b[7] = 15 - b[3] - b[5];
17 b[4] = 15 - b[1] - b[7];
18 b[6] = 15 - b[3] - b[9];
19 if(b[4]+b[5]+b[6]==15&&b[7]+b[8]+b[9]==15)
20 {
21 f = 0;
22 memset(a,0,sizeof(a));
23 for (int k=1;k<10;k++) a[b[k]]++;
24 for (int k=1;k<10;k++) if(a[k]<=0||a[k]>1){f = 1;break;}
25 if(f==0){
26 for (int k=1;k<10;k++)
27 {
28 cout<< b[k] <<' ';
29 if(k%3==0)cout << endl;
30 }
31 cout << endl;
32 }
33 }
34 }
35 }
36 }
37 system("pause");
38 return 0;
39 }
40
]]>
RSA is one of the most powerful methods to encrypt data. The RSA algorithm is described as follow:
> choose two large prime integer p, q
> calculate n = p × q, calculate F(n) = (p - 1) × (q - 1)
> choose an integer e(1 < e < F(n)), making gcd(e, F(n)) = 1, e will be the public key
> calculate d, making d × e mod F(n) = 1 mod F(n), and d will be the private key
You can encrypt data with this method :
C = E(m) = me mod n
When you want to decrypt data, use this method :
M = D(c) = cd mod n
Here, c is an integer ASCII value of a letter of cryptograph and m is an integer ASCII value of a letter of plain text.
Now given p, q, e and some cryptograph, your task is to "translate" the cryptograph into plain text.
> choose two large prime integer p, q
> calculate n = p × q, calculate F(n) = (p - 1) × (q - 1)
> choose an integer e(1 < e < F(n)), making gcd(e, F(n)) = 1, e will be the public key
> calculate d, making d × e mod F(n) = 1 mod F(n), and d will be the private key
You can encrypt data with this method :
C = E(m) = me mod n
When you want to decrypt data, use this method :
M = D(c) = cd mod n
Here, c is an integer ASCII value of a letter of cryptograph and m is an integer ASCII value of a letter of plain text.
Now given p, q, e and some cryptograph, your task is to "translate" the cryptograph into plain text.
Input
Each case will begin with four integers p, q, e, l followed by a line of cryptograph. The integers p, q, e, l will be in the range of 32-bit integer. The cryptograph consists of l integers separated by blanks.
Output
For each case, output the plain text in a single line. You may assume that the correct result of plain text are visual ASCII letters, you should output them as visualable letters with no blank between them.
Sample Input
101 103 7 11 7716 7746 7497 126 8486 4708 7746 623 7298 7357 3239
Sample Output
浠ヤ負瑕佹暟緇勶紝鎯蟲兂錛屼笉鏄偅涔堝洖浜嬶紒I-LOVE-ACM.
鎸虹畝鍗曠殑~~
1 #include<iostream>
2 using namespace std;
3 int main(){
4 int p,q,e,t;
5 while (cin>>p>>q>>e>>t)
6 {
7 int n = p*q,fn = (p-1)*(q-1);
8 int d =1;
9 while ((d*e)%fn!=1)d++;
10 int c;
11 for (int i=0;i<t;i++)
12 {
13 cin>>c;
14 int temp=1;
15 for (int j=1;j<=d;j++)
16 {
17 temp*=c;
18 temp%=n;
19 }
20 cout<<char(temp);
21 }
22 cout<<endl;
23 }
24 return 0;
25 }
2 using namespace std;
3 int main(){
4 int p,q,e,t;
5 while (cin>>p>>q>>e>>t)
6 {
7 int n = p*q,fn = (p-1)*(q-1);
8 int d =1;
9 while ((d*e)%fn!=1)d++;
10 int c;
11 for (int i=0;i<t;i++)
12 {
13 cin>>c;
14 int temp=1;
15 for (int j=1;j<=d;j++)
16 {
17 temp*=c;
18 temp%=n;
19 }
20 cout<<char(temp);
21 }
22 cout<<endl;
23 }
24 return 0;
25 }
]]>