銆愰瑙ｃ戯細鏈灝忔瘮鐜囩敓鎴愭爲鍜?1鍒嗘暟瑙勫垝閮芥槸浠庤繖閲屽鍒扮殑銆?br /> 鏈灝忔瘮鐜囩敓鎴愭爲鐨勬ā鍨嬪叾瀹炲氨鏄?1鍒嗘暟瑙勫垝錛屽叧浜?1鍒嗘暟瑙勫垝鐨勮祫鏂欙紝緗戜笂涓澶у爢銆?br /> 綆鍗曡涓嬪仛娉曪紝灝辨槸鏋勯犱竴涓?1鍒嗘暟瑙勫垝鐨勫紡瀛愶紝鐒跺悗鍙樺艦錛屽埄鐢∕ST姹傚彇褰撳墠鐨勬渶灝忓箋?br /> 鐒跺悗鍒╃敤浜屽垎鎴栬匘inkelbach閫艱繎鍒版渶浼樿В鍗充負絳旀銆?br />
銆愪唬鐮併戯細
1 #include "iostream"
2 #include "cstdio"
3 #include "cstring"
4 #include "algorithm"
5 #include "vector"
6 #include "queue"
7 #include "cmath"
8 #include "string"
9 #include "cctype"
10 #include "map"
11 #include "iomanip"
12 using namespace std;
13 #define pb push_back
14 #define lc(x) (x << 1)
15 #define rc(x) (x << 1 | 1)
16 #define lowbit(x) (x & (-x))
17 #define ll long long
18 #define maxn 1050
19 #define eps 1e-6
20
21 int n;
22 double d[maxn][maxn];
23 int cost[maxn][maxn];
24
25 struct Point {
26 double x, y;
27 int z;
28 }p[maxn];
29
30 double getdist(Point &a, Point &b) {
31 return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
32 }
33
34 int getcost(Point &a, Point &b) {
35 return abs(a.z - b.z);
36 }
37
38 double prim(int s, double &l) {
39 double anslen = 0.0, anscost = 0.0;
40 bool visit[maxn];
41 double dist[maxn], tmp, ans = 0.0;
42 int pre[maxn];
43 for(int i = 0; i < n; i++) visit[i] = false, dist[i] = 1e30;
44 dist[s] = 0.0, pre[s] = s;
45 while(1) {
46 int u = -1;
47 for(int i = 0; i < n; i++) {
48 if(!visit[i] && (u == -1 || dist[i] < dist[u])) u = i;
49 }
50 if(u == -1) break;
51 anscost += cost[pre[u]][u];
52 anslen += d[pre[u]][u];
53 ans += dist[u];
54 visit[u] = true;
55 for(int i = 0; i < n; i++) {
56 if(!visit[i] && (tmp = 1.0 * cost[u][i] - l * d[u][i]) < dist[i])
57 dist[i] = tmp, pre[i] = u;
58 }
59 }
60 l = anscost / anslen;
61 return ans;
62 }
63
64 void solve() {
65 double ans= 0.0, tmp = 0.0;
66 while(1) {
67 double tmp = prim(0, ans);
68 if(fabs(tmp) < eps) break;
69 }
70 printf("%.3f\n", ans);
71 }
72
73 int main() {
74 while(scanf("%d", &n) && n) {
75 for(int i = 0; i < n; i++) {
76 scanf("%lf%lf%d", &p[i].x, &p[i].y, &p[i].z);
77 for(int j = 0; j < i; j++) {
78 d[i][j] = d[j][i] = getdist(p[i], p[j]);
79 cost[i][j] = cost[j][i] = getcost(p[i], p[j]);
80 }
81 }
82 solve();
83 }
84 return 0;
85 }
86
2 #include "cstdio"
3 #include "cstring"
4 #include "algorithm"
5 #include "vector"
6 #include "queue"
7 #include "cmath"
8 #include "string"
9 #include "cctype"
10 #include "map"
11 #include "iomanip"
12 using namespace std;
13 #define pb push_back
14 #define lc(x) (x << 1)
15 #define rc(x) (x << 1 | 1)
16 #define lowbit(x) (x & (-x))
17 #define ll long long
18 #define maxn 1050
19 #define eps 1e-6
20
21 int n;
22 double d[maxn][maxn];
23 int cost[maxn][maxn];
24
25 struct Point {
26 double x, y;
27 int z;
28 }p[maxn];
29
30 double getdist(Point &a, Point &b) {
31 return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
32 }
33
34 int getcost(Point &a, Point &b) {
35 return abs(a.z - b.z);
36 }
37
38 double prim(int s, double &l) {
39 double anslen = 0.0, anscost = 0.0;
40 bool visit[maxn];
41 double dist[maxn], tmp, ans = 0.0;
42 int pre[maxn];
43 for(int i = 0; i < n; i++) visit[i] = false, dist[i] = 1e30;
44 dist[s] = 0.0, pre[s] = s;
45 while(1) {
46 int u = -1;
47 for(int i = 0; i < n; i++) {
48 if(!visit[i] && (u == -1 || dist[i] < dist[u])) u = i;
49 }
50 if(u == -1) break;
51 anscost += cost[pre[u]][u];
52 anslen += d[pre[u]][u];
53 ans += dist[u];
54 visit[u] = true;
55 for(int i = 0; i < n; i++) {
56 if(!visit[i] && (tmp = 1.0 * cost[u][i] - l * d[u][i]) < dist[i])
57 dist[i] = tmp, pre[i] = u;
58 }
59 }
60 l = anscost / anslen;
61 return ans;
62 }
63
64 void solve() {
65 double ans= 0.0, tmp = 0.0;
66 while(1) {
67 double tmp = prim(0, ans);
68 if(fabs(tmp) < eps) break;
69 }
70 printf("%.3f\n", ans);
71 }
72
73 int main() {
74 while(scanf("%d", &n) && n) {
75 for(int i = 0; i < n; i++) {
76 scanf("%lf%lf%d", &p[i].x, &p[i].y, &p[i].z);
77 for(int j = 0; j < i; j++) {
78 d[i][j] = d[j][i] = getdist(p[i], p[j]);
79 cost[i][j] = cost[j][i] = getcost(p[i], p[j]);
80 }
81 }
82 solve();
83 }
84 return 0;
85 }
86
y @ The Angry Teletubbies 2012-02-26 16:49 鍙戣〃璇勮
]]>銆愰瑙ｃ戯細濂芥槑鏄劇殑01鍒嗘暟瑙勫垝銆?br /> 璁鵑泦鍚坸[]錛寈[i]=0琛ㄧず涓嶉夌i涓猼est錛寈[i]=1琛ㄧず閫夌i涓猼est.
鏃㈡垜浠r = (∑a[i]*x[i]) / (∑b[i]*x[i]) 鏈澶э紝涓?#8721;x[i] = k.
鐒跺悗瑕佹瀯閫犲瓙闂錛岃緇嗙殑涓嶆兂鍐欎簡濂介夯鐑︺?br /> 鏈鍚庝笂浜屽垎鎴栬卍inkelbach鍗沖彲銆?br />
銆愪唬鐮併戯細
1 #include "iostream"
2 #include "cstdio"
3 #include "cstring"
4 #include "algorithm"
5 #include "vector"
6 #include "queue"
7 #include "cmath"
8 #include "string"
9 #include "cctype"
10 #include "map"
11 #include "iomanip"
12 using namespace std;
13 #define pb push_back
14 #define lc(x) (x << 1)
15 #define rc(x) (x << 1 | 1)
16 #define lowbit(x) (x & (-x))
17 #define ll long long
18 #define eps 1e-6
19 #define maxn 1050
20 int n, k;
21
22 struct Point {
23 int a, b;
24 double c;
25 bool operator <(const Point &x) const {
26 return c < x.c;
27 }
28 }p[maxn];
29
30 double work(double l) {
31 for(int i = 0; i < n; i++) p[i].c = 1.0 * p[i].a - l * p[i].b;
32 sort(p, p + n);
33 double sum = 0.0, suma = 0.0, sumb = 0.0;
34 for(int i = k; i < n; i++) sum += p[i].c, suma += p[i].a, sumb += p[i].b;
35 return suma / sumb;
36 }
37
38 void solve() {
39 double ans = 0.0, tmp = 0.0;
40 while(1) {
41 tmp = work(ans);
42 if(fabs(tmp - ans) < eps) break;
43 ans = tmp;
44 }
45 printf("%d\n", (int)(100.0 * (tmp + 0.005)));
46 }
47
48 int main() {
49 while(~scanf("%d%d", &n, &k)) {
50 if(!n && !k) break;
51 for(int i = 0; i < n; i++) scanf("%d", &p[i].a);
52 for(int i = 0; i < n; i++) scanf("%d", &p[i].b);
53 solve();
54 }
55 return 0;
56 }
57
2 #include "cstdio"
3 #include "cstring"
4 #include "algorithm"
5 #include "vector"
6 #include "queue"
7 #include "cmath"
8 #include "string"
9 #include "cctype"
10 #include "map"
11 #include "iomanip"
12 using namespace std;
13 #define pb push_back
14 #define lc(x) (x << 1)
15 #define rc(x) (x << 1 | 1)
16 #define lowbit(x) (x & (-x))
17 #define ll long long
18 #define eps 1e-6
19 #define maxn 1050
20 int n, k;
21
22 struct Point {
23 int a, b;
24 double c;
25 bool operator <(const Point &x) const {
26 return c < x.c;
27 }
28 }p[maxn];
29
30 double work(double l) {
31 for(int i = 0; i < n; i++) p[i].c = 1.0 * p[i].a - l * p[i].b;
32 sort(p, p + n);
33 double sum = 0.0, suma = 0.0, sumb = 0.0;
34 for(int i = k; i < n; i++) sum += p[i].c, suma += p[i].a, sumb += p[i].b;
35 return suma / sumb;
36 }
37
38 void solve() {
39 double ans = 0.0, tmp = 0.0;
40 while(1) {
41 tmp = work(ans);
42 if(fabs(tmp - ans) < eps) break;
43 ans = tmp;
44 }
45 printf("%d\n", (int)(100.0 * (tmp + 0.005)));
46 }
47
48 int main() {
49 while(~scanf("%d%d", &n, &k)) {
50 if(!n && !k) break;
51 for(int i = 0; i < n; i++) scanf("%d", &p[i].a);
52 for(int i = 0; i < n; i++) scanf("%d", &p[i].b);
53 solve();
54 }
55 return 0;
56 }
57
y @ The Angry Teletubbies 2012-02-19 15:38 鍙戣〃璇勮
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