题意：给一张有向带权图求出最小树形图。
思路：最小树形图模版题看了大概思想http://hi.baidu.com/bin183/item/5d93ef69ceb541176895e682
学习了邝巨巨的模版，然后抄了一遍模版。a掉了这道题。
代码如下：
1 /************************************************** 2 * Author : xiaohao Z 3 * Blog : http://www.cnblogs.com/shu-xiaohao/ 4 * Last modified : 2014-02-06 21:32 5 * Filename : uva_11183.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 double eps = 1E-6; 33 const int LEN = 1010; 34 int Map[LEN][LEN]; 35 int pre[LEN], id[LEN], vis[LEN], in[LEN]; 36 37 struct E{ 38 int fr, to, val; 39 }; 40 41 int zhuliu(int root, int n, int m, E edge[]){ 42 int ret = 0, u, v; 43 while(1){ 44 for(int i=0; i<n; i++) in[i] = INF; 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] == INF) return INF; 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 main() 84 { 85 // freopen("in.txt", "r", stdin); 86 87 int a, b, val, T, n, m; 88 E edge[40010]; 89 scanf("%d", &T); 90 for(int kase = 1; kase <= T; kase++){ 91 memset(Map, INF, sizeof Map); 92 scanf("%d%d", &n, &m); 93 for(int i=0; i<m; i++){ 94 scanf("%d%d%d", &a, &b, &val); 95 Map[a][b] = min(Map[a][b], val); 96 } 97 int top = 0; 98 for(int i=0; i<n; i++) 99 for(int j=0; j<n; j++) 100 if(Map[i][j] != INF){ 101 edge[top].fr = i; 102 edge[top].to = j; 103 edge[top++].val = Map[i][j]; 104 } 105 int ans = zhuliu(0, n, m, edge); 106 printf("Case #%d: ", kase); 107 if(ans == INF) printf("Possums! "); 108 else printf("%d ", ans); 109 } 110 return 0; 111 }