题意：给出一颗树和树根，每条树边都有边权，要你砍断一些边，使得所有的叶子节点都与根节点分离，且要求砍断的边权之和最小。
以树根root为源点S，添加一个汇点T，对所有的非跟叶子节点i，加边(i,T,inf)，求最小割即可。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #define INF 1<<30 6 #define maxn 1010 7 #define maxm 10000 8 using namespace std; 9 10 int u[maxm],v[maxm],next[maxm],w[maxm]; 11 int first[maxn],d[maxn],work[maxn],q[maxn],vis[maxn]; 12 int e,S,T; 13 int degree[maxn]; 14 15 void init(){ 16 e = 0; 17 memset(first,-1,sizeof(first)); 18 } 19 20 void add_edge(int a,int b,int c){ 21 u[e] = a;v[e] = b;next[e] = first[a];w[e] = c;first[a] = e++; 22 u[e] = b;v[e] = a;next[e] = first[b];w[e] = 0;first[b] = e++; 23 } 24 25 int bfs(){ 26 int rear = 0; 27 memset(d,-1,sizeof(d)); 28 d[S] = 0;q[rear++] = S; 29 for(int i = 0;i < rear;i++){ 30 for(int j = first[q[i]];j != -1;j = next[j]) 31 if(w[j] && d[v[j]] == -1){ 32 d[v[j]] = d[q[i]] + 1; 33 q[rear++] = v[j]; 34 if(v[j] == T) return 1; 35 } 36 } 37 return 0; 38 } 39 40 int dfs(int cur,int a){ 41 if(cur == T) return a; 42 for(int &i = work[cur];i != -1;i = next[i]){ 43 if(w[i] && d[v[i]] == d[cur] + 1) 44 if(int t = dfs(v[i],min(a,w[i]))){ 45 w[i] -= t;w[i^1] += t; 46 return t; 47 } 48 } 49 return 0; 50 } 51 52 int dinic(){ 53 int ans = 0; 54 while(bfs()){ 55 memcpy(work,first,sizeof(first)); 56 while(int t = dfs(S,INF)) ans += t; 57 } 58 return ans; 59 } 60 61 int main() 62 { 63 int n,root; 64 while(scanf("%d%d",&n,&root) == 2){ 65 if(!n && !root) break; 66 memset(degree,0,sizeof(degree)); 67 S = root,T = n+1; 68 init(); 69 for(int i = 0;i < n-1;i++){ 70 int a,b,c; 71 scanf("%d%d%d",&a,&b,&c); 72 add_edge(a,b,c); 73 add_edge(b,a,c); 74 degree[a]++;degree[b]++; 75 } 76 for(int i = 1;i <= n;i++){ 77 if(degree[i] == 1 && i != root) 78 add_edge(i,T,INF); 79 } 80 printf("%d ",dinic()); 81 } 82 return 0; 83 }