【luogu P2194 HXY烧情侣】 题解

题目链接：https://www.luogu.org/problemnew/show/P2194

第一问：缩点并且统计其强连通分量里的最小耗费。把所有强连通分量的最小耗费加起来。

第二问：统计在每个强连通分量里与最小耗费相同的点数。乘法原理统计所有强连通分量答案。

#include <stack>

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

const int maxn = 300000 + 10;

const int inf = 0x7fffffff;

const int mod = 1e9 + 7;

struct edge{

	int from, to, next;

}e[maxn<<2];

int head[maxn], cnt;

int n, m, ans1, ans2 = 1, dfn[maxn], low[maxn], tim, color[maxn], num, val[maxn], minpay[maxn], tot[maxn];

bool vis[maxn];

stack<int> s;

void add(int u, int v)

{

	e[++cnt].from = u;

	e[cnt].next = head[u];

	e[cnt].to = v;

	head[u] = cnt;

}

void tarjan(int x)

{

	dfn[x] = low[x] = ++tim;

	vis[x] = 1; s.push(x);

	for(int i = head[x]; i != -1; i = e[i].next)

	{

		int v = e[i].to;

		if(!dfn[v])

		{

			tarjan(v);

			low[x] = min(low[x], low[v]);

		}

		else if(vis[v])

		{

			low[x] = min(low[x], low[v]);

		}

	}

	if(dfn[x] == low[x])

	{

		color[x] = ++num;

		vis[x] = 0;

		minpay[num] = min(minpay[num], val[x]);

		while(s.top() != x)

		{

			color[s.top()] = num;

			vis[s.top()] = 0;

			minpay[num] = min(minpay[num], val[s.top()]);

			s.pop();

		}

		s.pop();

	}

}

int main()

{

	memset(head, -1, sizeof(head));

	scanf("%d",&n);

	for(int i = 1; i <= n; i++) minpay[i] = inf;

	for(int i = 1; i <= n; i++) scanf("%d",&val[i]);

	scanf("%d",&m);

	for(int i = 1; i <= m; i++)

	{

		int u, v;

		scanf("%d%d",&u,&v);

		add(u,v);

	}

	for(int i = 1; i <= n; i++)

		if(!dfn[i]) tarjan(i);

	//for(int i = 1; i <= n; i++) cout<<minpay[i];

	for(int i = 1; i <= num; i++) ans1 += minpay[i];

	cout<<ans1<<" ";

	for(int i = 1; i <= n; i++)

	{

		if(val[i] == minpay[color[i]])

		tot[color[i]]++;

	}

	for(int i = 1; i <= num; i++) ans2 = (ans2*tot[i])%mod;

	cout<<ans2<<endl;

	return 0;

}