HDU 1827：Summer Holiday（强连通）

http://acm.hdu.edu.cn/showproblem.php?pid=1827

思路：强连通分量缩点后找入度为0的点，然后对于属于该强连通分量的找一个最小耗费的入口。

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <iostream>
 #include <queue>
 #include <cmath>
 #include <map>
 #include <vector>
 #include <stack>
 using namespace std;
 #define N 80010
 #define M 30010
 #define INF 0x3f3f3f3f
 struct node
 {
     int u, v, next;
 }edge[N*];
 stack<int> sta;
 int tot, cnt, num, head[N], belong[N], dfn[N], low[N], n, m, a[N], in[N], mi[N];
 bool vis[N];

 void init()
 {
     tot = cnt = num = ;
     while(sta.size()) sta.pop();
     memset(head, -, sizeof(head));
     memset(belong, , sizeof(belong));
     memset(dfn, , sizeof(dfn));
     memset(low, , sizeof(low));
     memset(vis, false, sizeof(vis));
     memset(in, , sizeof(in));
     memset(mi, INF, sizeof(mi));
 }

 void add(int u, int v)
 {
     edge[tot].u = u; edge[tot].v = v; edge[tot].next = head[u]; head[u] = tot++;
 }

 void tarjan(int u)
 {
     vis[u] = ;
     sta.push(u);
     dfn[u] = low[u] = ++cnt;
     for(int i = head[u]; ~i; i = edge[i].next) {
         int v = edge[i].v;
         if(!dfn[v]) {
             tarjan(v);
             if(low[v] < low[u]) low[u] = low[v];
         } else if(vis[v]) {
             if(dfn[v] < low[u]) low[u] = dfn[v];
         }
     }
     if(dfn[u] == low[u]) {
         ++num;
         int top = -;
         while(top != u) {
             top = sta.top(); sta.pop();
             belong[top] = num;
             vis[top] = ;
         }
     }
 }

 int main()
 {
     while(~scanf("%d%d", &n, &m)) {
         init();
         for(int i = ; i <= n; i++) scanf("%d", &a[i]);
         for(int i = ; i < m; i++) {
             int u, v;
             scanf("%d%d", &u, &v);
             add(u, v);
         }
         for(int i = ; i <= n; i++)
             if(!dfn[i]) tarjan(i);
         for(int u = ; u <= n; u++) {
             for(int i = head[u]; ~i; i = edge[i].next) {
                 int v = edge[i].v;
                 if(belong[u] != belong[v]) { //缩点后找入度为0的点
                     in[belong[v]]++;
                 }
             }
         }
         int ans = , sum = ;
         for(int i = ; i <= num; i++) {
             if(!in[i]) {
                 ans++;
                 for(int j = ; j <= n; j++) {
                     if(belong[j] == i) {
                         if(a[j] < mi[i]) {
                             mi[i] = a[j]; //对于不同的强连通分量之间找耗费最小的入口
                         }
                     }
                 }
                 sum += mi[i];
             }
         }
         printf("%d %d\n", ans, sum);
     }
     return ;
 }

 /*
 4 3
 1 2 3 4
 1 2
 1 3
 1 4
 */