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hdoj 4253 Two Famous Companies - 二分 + 最��生成树

y @ The Angry Teletubbies — Tue, 17 Jul 2012 17:36:00 GMT

【题意】：�l�出n个点�Q�m条边�Q�边分�ؓ两种�Q�一�U�是A公司的，一�U�是B公司的。边上有权��|��问用n-1条边把n个点�q��v来的最��费用是多少�Q�其中A公司的边刚好有k条。题目保证有解�?br />
【题解】：很明昄��到是求一颗最��生成树�Q�不�q�有一个限制就是刚刚好有k条边是A公司的。想了很久不会做�Q�看别�h代码的。二分出一个最大值delta使得A公司的边加上�q�个值后再求MST时A公司的边有大于等于k条，然后�{�案��是cost of MST - k * delta。思想其实是加上一个delta值去��D��{�案�Q�最后可以求��L��MST�Q�如果最后求出的MST的A公司的边多于k条，一定存在与A公司�{�效且等��L��B公司边，替换�q�来卛_��?br />
【代码】：

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 #include "set"
13 #include "utility"
14 using namespace std;
15 typedef pair<int, int> pii;
16 #define pb push_back
17 #define mp make_pair
18 #define fi first
19 #define se second
20 #define sof(x) sizeof(x)
21 #define lc(x) (x << 1)
22 #define rc(x) (x << 1 | 1)
23 #define lowbit(x) (x & (-x))
24 #define ll long long
25 struct Edge {
26     int u, v, w, id;
27     Edge(){}
28     Edge(int _u, int _v, int _w, int _id) {
29         u = _u, v = _v, w = _w, id = _id;
30     }
31     bool operator<(const Edge &x) const {
32         if(w != x.w) return w < x.w;
33         else return id < x.id;
34     }
35 }et[2][100050], e;
36 int tot, tot1;
37 int n, m, k, cost;
38 int fa[50050];
39
40 int find(int x) {
41     return (x == fa[x]) ? x : fa[x] = find(fa[x]);
42 }
43
44 bool merge(int u, int v) {
45     u = find(fa[u]), v = find(fa[v]);
46     if(u != v) {
47         fa[u] = v;
48         return true;
49     } else return false;
50 }
51
52 bool check(int w) {
53     cost = 0;
54     int cnt = 0;
55     for(int i = 0; i < n; i++) fa[i] = i;
56     int i = 0, j = 0;
57     while(i < tot || j < tot1) {
58         if(et[0][i].w + w <= et[1][j].w) {
59             e = et[0][i++];
60             e.w += w;
61         } else e = et[1][j++];
62         if(merge(e.u, e.v)) {
63             if(!e.id) cnt++;
64             cost += e.w;
65         }
66     }
67     return cnt >= k;
68 }
69
70 int main() {
71     int Case = 1;
72     while(~scanf("%d%d%d", &n, &m, &k)) {
73         tot = tot1 = 0;
74         for(int i = 0; i < m; i++) {
75             int u, v, w, id;
76             scanf("%d%d%d%d", &u, &v, &w, &id);
77             if(id) et[1][tot1++] = Edge(u, v, w, id);
78             else et[0][tot++] = Edge(u, v, w, id);
79         }
80         sort(et[0], et[0] + tot);
81         sort(et[1], et[1] + tot1);
82         et[0][tot].w = et[1][tot1].w = 1 << 30;
83         int l = -100, r = 100;
84         int w;
85         while(l <= r) {
86             int mid = (l + r) / 2;
87             if(check(mid)) w = mid, l = mid + 1;
88             else r = mid - 1;
89         }
90         check(w);
91         printf("Case %d: %d\n", Case++, cost - w * k);
92     }
93     return 0;
94 }
95

y @ The Angry Teletubbies 2012-07-18 01:36 发表评论

hdoj 3586 Information Disturbing - 二分 + 树dp

y @ The Angry Teletubbies — Thu, 05 Apr 2012 06:10:00 GMT

【题意】：�l�出一��|��Q�根节点�?�Q�每条边有一个代仗��要求切��d��所有的叶子节点�Q�且��M��价不能超�q�m�Q�问切去的边中的最大的�Ҏ��是多少�?/span>

【题解】：最大值最��问题，基本上就是二分答�?判定�?br /> �q�题的判定需要用到树dp�?br /> 讄��态dp[u]��Z��u为根的子树切��d��下方叶子节点的最��代仗��?br /> 非叶子节点：
dp[u] += min(dp[v], w(u, v)), v �?u 儿子�?br /> 叶子节点�Q?br /> dp[u] = inf;
最后判断dp[1]与m的关�p�d��可以了�?br />
【代码】：

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 mp make_pair
15 #define fi first
16 #define se second
17 #define lc(x) (x << 1)
18 #define rc(x) (x << 1 | 1)
19 #define lowbit(x) (x & (-x))
20 #define ll long long
21 #define maxn 1010
22 const int inf = 1 << 20;
23 int n, m, limit;
24 int dp[maxn];
25 struct Edge {
26     int v, w;
27     Edge(){}
28     Edge(int _v, int _w) {
29         v = _v, w = _w;
30     }
31 };
32 vector vec[maxn];
33
34 void dfs(int u, int fa) {
35     dp[u] = 0;
36     int size = vec[u].size();
37     for(int i = 0; i < size; i++) {
38         int v = vec[u][i].v, w = vec[u][i].w;
39         if(v == fa) continue;
40         if(w > limit) w = inf;
41         dfs(v, u);
42         dp[u] += min(dp[v], w);
43     }
44     if(!dp[u]) dp[u] = inf;
45 }
46
47 void solve() {
48     int ans = -1;
49     int l = 1, r = 1000;
50     while(l <= r) {
51         int mid = (l + r) >> 1;
52         limit = mid;
53         dfs(1, -1);
54         if(dp[1] <= m) ans = mid, r = mid - 1;
55         else l = mid + 1;
56     }
57     cout << ans << endl;
58 }
59
60 int main() {
61     while(~scanf("%d%d", &n, &m)) {
62         if(!n) break;
63         for(int i = 1; i <= n; i++) vec[i].clear();
64         for(int i = 1; i < n; i++) {
65             int u, v, w;
66             scanf("%d%d%d", &u, &v, &w);
67             vec[u].pb(Edge(v, w));
68             vec[v].pb(Edge(u, w));
69         }
70         solve();
71     }
72     return 0;
73 }
74

y @ The Angry Teletubbies 2012-04-05 14:10 发表评论

y @ The Angry Teletubbies — Wed, 08 Feb 2012 16:10:00 GMT

【题意】：�l�出n件刚�z�完的衣服，每�g衣服有一个属性ai�Q�表�C�改衣服在自焉��q�的条�g需要ti分钟。现在有一台烘�q�机�Q�每分钟你可以选择把�Q意一件衣服放�q�去�Q�那么这件衣服风�q�的旉��减��k分钟�Q�而每分钟对于不在烘干机的衣服�Q�风�q�时间都减少1。问最��用多少旉��使所有衣服都�q�了�?br />
【题解】：直接二分�{�案�Q�然后判断答案的正确性�?br /> 假设当前二分的答案�ؓ t�Q�那么：
对于ai <= t的衣服，昄��让它们自焉��q�就可以了�?br /> 对于ai > t的衣服，我们需要知道该衣服最��用多少�ơ烘�q�机�?br /> 设该衣服用了x1分钟风干�Q�用了x2分钟烘干机�?br /> 那么�?x1 + x2 = t �?ai <= x1 + x2 * k�Q�联立两式可�?x2 >= (ai - t) / (k - 1)�Q�即最��用次��Cؓ[(ai - t) / (k - 1)] 的最��上界�?br /> 最后，判断一下��M��用次数是否少�?t 卛_��?br />
【代码】：

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 100050
19 int n;
20 ll val[maxn], ans, k;
21
22 bool check(ll t) {
23     ll cnt = 0;
24     for(int i = 0; i < n; i++) {
25         if(val[i] > t) {
26             cnt += (val[i] - t + k - 2) / (k - 1);
27             if(cnt > t) return false;
28         }
29     }
30     return true;
31 }
32
33 void solve() {
34     ll l = 1, r = ans;
35     while(l <= r) {
36         ll mid = (l + r) >> 1;
37         if(check(mid)) ans = mid, r = mid - 1;
38         else l = mid + 1;
39     }
40     printf("%lld\n", ans);
41 }
42
43 int main() {
44     while(~scanf("%d", &n)) {
45         ans = 0;
46         for(int i = 0; i < n; i++) {
47             scanf("%lld", &val[i]);
48             ans = max(ans, val[i]);
49         }
50         scanf("%lld", &k);
51         if(k == 1) printf("%lld\n", ans);
52         else solve();
53     }
54     return 0;
55 }
56

y @ The Angry Teletubbies 2012-02-09 00:10 发表评论

y @ The Angry Teletubbies — Fri, 11 Nov 2011 10:59:00 GMT

【题意】：在三�l�空间中�Q�给出n个点(x,y,z) (z>=0)�Q�要求用一个体�U�最��的圆锥包围所有的点，输出该圆锥的高和半径�?br />
【题解】：可以先分析一下，当这个圆锥的高很大时�Q�圆锥的体积非常大；当这个圆锥的高太��的时候，圆锥的半径就会非常大�Q�此时圆锥的体积�q�是很大�?br /> 于是我们可以得出一个结论，圆锥的高不能太小也不能太大�?br /> �q�里有一步预处理�Q�就是把�I�间直角坐标�p��{换�ؓ柱面坐标�p�R�?x,y,z) => (sqrt(x*x+y*y),z);
有了上面的结论，我们��可以利用三分极值来��定�q�个高的最优��|��然后通过�怼�三角形就可以求出圆锥的半径�?br /> 此题也可以通过三分半径来求解�?br />
【代码】：

1 #include "iostream"
2 #include "cstdio"
3 #include "cstring"
4 #include "cmath"
5 using namespace std;
6 #define maxn 10500
7 #define eps 1e-8
8 #define pi 3.1415926535897932384626433832795
9 int n;
10 double rr;
11 struct Point {
12     double x, y, r, z;
13     Point(){}
14     Point(double _x, double _y, double _z) {
15         x = _x, y = _y, z = _z, r = sqrt(_x * _x + _y * _y) ;
16     }
17 }point[maxn];
18
19 double work(double h) {
20     double tmp;
21     rr = 0.0;
22     for(int i = 0; i < n; i++) {
23         double r = point[i].r, z = point[i].z;
24         double tmp = h * r / (h - z);
25         rr = max(rr, tmp);
26     }
27     double area = pi * rr * rr * h / 3.0;
28     return area;
29 }
30
31 void solve(double res) {
32     double low = res, high = 10000000, mid, mmid, ansh, ansr;
33     while(low + eps < high) {
34         mid = (low + high) / 2.0;
35         mmid = (mid + high) / 2.0;
36         if(work(mid) < work(mmid)) {
37             high = mmid;
38             ansh = mid;
39         } else {
40             low = mid;
41             ansh = mmid;
42         }
43         ansr = rr;
44     }
45     printf("%.3f %.3f\n", ansh, ansr);
46
47 }
48
49 int main() {
50     int T;
51     double x, y, z;
52     scanf("%d", &T);
53     while(T--) {
54         scanf("%d", &n);
55         double low = 0.0;
56         for(int i = 0; i < n; i++) {
57             scanf("%lf%lf%lf", &x, &y, &z);
58             point[i] = Point(x, y, z);
59             low = max(low, z);
60         }
61         solve(low);
62     }
63     return 0;
64 }

y @ The Angry Teletubbies 2011-11-11 18:59 发表评论

理论片一区二区在线,欧美国产在线电影,国产精品国产三级国产a

hdoj 4253 Two Famous Companies - 二分 + 最���生成树

hdoj 3586 Information Disturbing - 二分 + 树dp

hdoj 4253 Two Famous Companies - 二分 + 最��生成树