codeforces 427C tarjan template title

Meaning of the questions:

N give you points for each point of the construction cost of the police station

For this chart, the police station and each point in the same strong component of China Unicom can be in the care of the police station to ask you a minimum cost and the current cost of the program number

Ideas:

tarjan template title ~~~ (Special thanks to Lei brother's portrait taught ლ (° ◕ `ƹ'◕ ლ) )

AC Code:

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>

using namespace std;
const int MaxN = 1e5 + 5;
typedef long long LL;
const LL mod = 1e9 + 7;

int w[MaxN],stack[MaxN],dfn[MaxN],low[MaxN],col[MaxN];
int n,m,cnt = 0,Top = 0,haha = 0;
bool vis[MaxN];
vector<int> G[MaxN],E[MaxN];

void tarjan(int u){
	dfn[u] = ++cnt;
	low[u] = dfn[u];
	stack[++Top] = u;
	vis[u] = 1;//在stack里
	int len = G[u].size();
	for(int i = 0;i < len; i++){
		int v = G[u][i];
		if(!dfn[v]){//没访问过
			tarjan(v);
			low[u] = min(low[u],low[v]);
		}
		else if(vis[v]){//访问过 && 在栈里
			low[u] = min(low[u],dfn[v]);
		}
	}
	if(low[u] == dfn[u]){
		haha++;
		while(1){
			int cur = stack[Top];
			vis[cur] = 0;
			Top--;
			col[cur] = haha;
			if(cur == u) break;
		}
	}
}

int main()
{
	scanf("%d",&n);
	for(int i = 1;i <= n; i++) scanf("%d",&w[i]);
	scanf("%d",&m);
	for(int i = 1;i <= m; i++){
		int u,v;
		scanf("%d %d",&u,&v);
		G[u].push_back(v);
	}
	for(int i = 1;i <= n; i++){
		if(!dfn[i]) tarjan(i);
	}//搜索所有强联通分量
	for(int i = 1;i <= n; i++){
		if(!E[col[i]].size()) E[col[i]].push_back(i);
		else if(w[i] < w[E[col[i]][0]]){
			E[col[i]].clear();
			E[col[i]].push_back(i);
		}
		else if(w[i] == w[E[col[i]][0]]){
			E[col[i]].push_back(i);
		}
	}
	LL cost = 0LL,ways = 1LL;
	for(int i = 1;i <= haha; i++){
		cost += w[E[i][0]];
		ways = (ways * (LL)E[i].size()) % mod;
	}
	printf("%lld %lld\n",cost,ways);
	return 0;
}