题目：http://vjudge.net/problem/UVA-11865
题意：以0为根构建树形图，其中每条边有两个权值，一个为花费一个为流量，要求构建树形图时花费不超过C且使构建树形图所用到的边中流量的最小值最大
思路：构建树形图这个简单，套模板即可，怎么使流量的最小值最大化呢？二分枚举流量的值，流量小于枚举值的边弃用，然后构建树形图，如果无法构建树形图或者花费超过了C，那么就向下调整枚举值，否则向上调整。注意在求最小树形图的过程中储存图的数组会被改变，因此用一个图的副本来求最小树形图
#include <iostream> #include <cstdio> #include <cstring> #include <string> #include <algorithm> #include <cmath> #include <queue> #include <map> using namespace std; const int N = 110, INF = 0x3f3f3f3f; struct edge { int v, u, cap, cost; }mpa[N*N], g[N*N]; int n, m, c; int in[N], pre[N], id[N], vis[N]; int ma(int s, int n, int m, int mid) { int res = 0; while(true) { for(int i = 0; i < n; i++) in[i] = INF; for(int i = 0; i < m; i++) { int v = g[i].v, u = g[i].u; if(v != u && g[i].cap >= mid && g[i].cost < in[u]) in[u] = g[i].cost, pre[u] = v; } for(int i = 0; i < n; i++) { if(i == s) continue; if(in[i] == INF) return -1; } int num = 0; memset(id, -1, sizeof id); memset(vis, -1, sizeof vis); in[s] = 0; for(int i = 0; i < n; i++) { res += in[i]; int v = i; while(vis[v] != i && id[v] == -1 && v != s) vis[v] = i, v = pre[v]; if(v != s && id[v] == -1) { for(int j = pre[v]; j != v; j = pre[j]) id[j] = num; id[v] = num++; } } if(num == 0) break; for(int i = 0; i < n; i++) if(id[i] == -1) id[i] = num++; for(int i = 0; i < m; i++) { int v = g[i].v, u = g[i].u; g[i].v = id[v], g[i].u = id[u]; if(id[v] != id[u]) g[i].cost -= in[u]; } n = num, s = id[s]; } return res; } int main() { int t; scanf("%d", &t); while(t--) { scanf("%d%d%d", &n, &m, &c); int maxx = 0; for(int i = 0; i < m; i++) scanf("%d%d%d%d", &mpa[i].v, &mpa[i].u, &mpa[i].cap, &mpa[i].cost), maxx = max(maxx, mpa[i].cap); int l = 0, r = maxx, res = -1; while(l <= r) { int mid = (l + r) / 2; for(int i = 0; i < m; i++) g[i] = mpa[i];//复制图 int tmp = ma(0, n, m, mid); if(tmp == -1 || tmp > c) r = mid - 1;//无法构建树形图或者花费超过了C else l = mid + 1, res = mid; } if(res == -1) printf("streaming not possible.\n"); else printf("%d kbps\n", res); } return 0; }