中文题一道，好题啊，读题简单多了，做起来也不难，就是缩点，缩点完成后，入度为0的点 就是需要主人公特意需要去打电话的点，缩点后 对于某一个团的最小话费就很好求了，直接暴力查找就可以了，

#include<iostream>
#include<cstdio>
#include<list>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<cmath>
#include<memory.h>
#include<set>

#define ll long long

#define eps 1e-7

#define inf 0xfffffff
const ll INF = 1ll<<61;

using namespace std;

//vector<pair<int,int> > G;
//typedef pair<int,int > P;
//vector<pair<int,int> > ::iterator iter;
//
//map<ll,int >mp;
//map<ll,int >::iterator p;
//


typedef struct Node {
	int from,to;
	int nex;
};

Node edge[1000 * 10];

int head[2000 + 10],Stack[2000 + 10],low[2000 + 10],dfn[2000 + 10],id[2000 + 10],w[1000 + 10];

bool vis[2000 + 10];

int tot,vis_num,stack_num,scc_num;

void clear() {
	memset(head,-1,sizeof(head));
	memset(Stack,0,sizeof(Stack));
	memset(low,0,sizeof(low));
	memset(dfn,-1,sizeof(dfn));
	memset(id,0,sizeof(id));
	memset(vis,false,sizeof(vis));
	memset(w,0,sizeof(w));
	memset(edge,0,sizeof(edge));
	tot = 0;
	vis_num = 0;
	stack_num = 0;
	scc_num = 0;
}

void addedge(int u,int v) {
	edge[tot].from = u;
	edge[tot].to = v;
	edge[tot].nex = head[u];
	head[u] = tot++;
}

void tarjan(int v) {
	dfn[v] = low[v] = ++vis_num;
	vis[v] = true;
	Stack[stack_num++] = v;
	for(int i=head[v];i!=-1;i=edge[i].nex) {
		int u = edge[i].to;
		if(dfn[u] == -1) {
			tarjan(u);
			low[v] = min(low[u],low[v]);
		}
		else if(vis[u])
			low[v] = min(low[v],dfn[u]);
	}
	int tmp;
	if(low[v] == dfn[v]) {
		scc_num++;
		do {
			tmp = Stack[--stack_num];
			id[tmp] = scc_num;
			vis[tmp] = false;
		}while(tmp != v);
	}
}

void cal(int n) {
	for(int i=1;i<=n;i++)
		if(dfn[i] == -1)
			tarjan(i);
}

int main() {
	int n,m;
	while(scanf("%d %d",&n,&m) == 2) {
		clear();
		for(int i=1;i<=n;i++)
			scanf("%d",&w[i]);
		int a,b;
		while(m--) {
			scanf("%d %d",&a,&b);
			addedge(a,b);
		}
		cal(n);
		int in[2000 + 10];//入度
		int hehe[2000 + 10];//话费
		memset(in,0,sizeof(in));
		for(int i=0;i<tot;i++) {
			int u = edge[i].from,v = edge[i].to;
			if(id[u] != id[v])
				in[ id[v] ]++;
		}
		int ans1 = 0;
		for(int i=1;i<=scc_num;i++) {
			if(in[i] == 0)
				ans1++;
			hehe[i] = inf;
		}
		for(int i=1;i<=n;i++) {
			int tmp = id[i];
			if(in[tmp] == 0)
				hehe[tmp] = min(hehe[tmp],w[i]);
		}
		int ans2 = 0;
		for(int i=1;i<=scc_num;i++)
			if(hehe[i] != inf)
				ans2 += hehe[i];
		printf("%d %d\n",ans1,ans2);
	}
	return	EXIT_SUCCESS;
}

HDU1827 Summer Holiday 强连通 Tarjan 缩点 统计