题意：有向带权图中每一条边上有一个带宽和长度，你的任务是在图中找出树形图他的费用小于cost，且是树中最小带宽最大。
思路：二分带宽+最小树形图。由于满足单调性（带宽范围越大花费越少）所以我们可以二分带宽然后求满足要求的边的最小树形图即可。
代码如下：
1 /************************************************** 2 * Author : xiaohao Z 3 * Blog : http://www.cnblogs.com/shu-xiaohao/ 4 * Last modified : 2014-02-07 13:53 5 * Filename : uva_11865.cpp 6 * Description : 7 * ************************************************/ 8 9 #include <iostream> 10 #include <cstdio> 11 #include <cstring> 12 #include <cstdlib> 13 #include <cmath> 14 #include <algorithm> 15 #include <queue> 16 #include <stack> 17 #include <vector> 18 #include <set> 19 #include <map> 20 #define MP(a, b) make_pair(a, b) 21 #define PB(a) push_back(a) 22 23 using namespace std; 24 typedef long long ll; 25 typedef pair<int, int> pii; 26 typedef pair<unsigned int,unsigned int> puu; 27 typedef pair<int, double> pid; 28 typedef pair<ll, int> pli; 29 typedef pair<int, ll> pil; 30 31 const int INF = 0x3f3f3f3f; 32 const ll LINF = 0x3f3f3f3f3f3f3f3fLL; 33 const double eps = 1E-6; 34 const int LEN = 201; 35 int pre[LEN], id[LEN], vis[LEN]; 36 ll in[LEN]; 37 struct E{ int fr, to, bmp;ll val;}; 38 E te[LEN*LEN]; 39 40 ll zhuliu(int root, int n, int m, E edge[]){ 41 ll ret = 0; 42 int u, v; 43 while(1){ 44 for(int i=0; i<n; i++) in[i] = LINF; 45 for(int i=0; i<m; i++) 46 if(edge[i].fr != edge[i].to && edge[i].val < in[edge[i].to]){ 47 pre[edge[i].to] = edge[i].fr; 48 in[edge[i].to] = edge[i].val; 49 } 50 for(int i=0; i<n; i++) 51 if(i != root && in[i] == LINF) return LINF; 52 int tn = 0; 53 memset(id, -1, sizeof id); 54 memset(vis, -1, sizeof vis); 55 in[root] = 0; 56 for(int i=0; i<n; i++){ 57 ret += in[i]; 58 v = i; 59 while(vis[v] != i && id[v] == -1 && v != root){ 60 vis[v] = i; 61 v = pre[v]; 62 } 63 if(v != root && id[v] == -1){ 64 for(int u=pre[v]; u!=v; u=pre[u])id[u] = tn; 65 id[v] = tn++; 66 } 67 } 68 if(tn == 0) break; 69 for(int i=0; i<n; i++) if(id[i] == -1)id[i] = tn++; 70 for(int i=0; i<m; ){ 71 v = edge[i].to; 72 edge[i].fr = id[edge[i].fr]; 73 edge[i].to = id[edge[i].to]; 74 if(edge[i].fr != edge[i].to) edge[i++].val -= in[v]; 75 else swap(edge[i], edge[--m]); 76 } 77 n = tn; 78 root = id[root]; 79 } 80 return ret; 81 } 82 83 int Makedge(int tbad, E edge[], int m){ 84 int top = 0; 85 for(int i=0; i<m; i++) 86 if(te[i].bmp >= tbad){ 87 edge[top].fr = te[i].fr; 88 edge[top].to = te[i].to; 89 edge[top++].val = te[i].val; 90 } 91 return top; 92 } 93 94 int main() 95 { 96 // freopen("in.txt", "r", stdin); 97 98 int T, n, m, cost; 99 E edge[LEN*LEN]; 100 scanf("%d", &T); 101 while(T--){ 102 int l = INF, r = 0; 103 scanf("%d%d%d", &n, &m, &cost); 104 for(int i=0; i<m; i++){ 105 scanf("%d%d%d%lld", &te[i].fr, &te[i].to, &te[i].bmp, &te[i].val); 106 l = min(l, te[i].bmp); 107 r = max(r, te[i].bmp); 108 } 109 Makedge(0, edge, m); 110 if(zhuliu(0, n, m, edge) > cost) printf("streaming not possible. "); 111 else { 112 while(l < r){ 113 int mid = (l+r+1)/2, tm = Makedge(mid, edge, m); 114 ll tans = zhuliu(0, n, tm, edge); 115 if(tans <= cost) l = mid; 116 else r = mid-1; 117 } 118 printf("%d kbps ", l); 119 } 120 } 121 return 0; 122 }