【luogu P1262 间谍网络】 题解

题目链接：https://www.luogu.org/problemnew/show/P1262
注意：
1.缩点时计算出入度是在缩完点的图上用color计算。不要在原来的点上计算。
2.枚举出入度时是在缩完点的图上计算。枚举范围到num。

#include <stack>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 10000 + 10;
struct edge{
	int from, next, to, len;
}e[maxn<<2];
int head[maxn], cnt;
int dfn[maxn], low[maxn], tim, color[maxn], num, rudu[maxn];
stack<int> s;
bool vis[maxn], okbuy[maxn], flag[maxn];
int n, p, m, money[maxn], minpay[maxn], ans;
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;
	s.push(x); vis[x] = 1;
	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(low[x] == dfn[x])
	{
		color[x] = ++num;
		vis[x] = 0;
		minpay[num] = min(minpay[num], money[x]);
		while(s.top() != x)
		{
			color[s.top()] = num;
			vis[s.top()] = 0;
			minpay[num] = min(minpay[num], money[s.top()]);
			s.pop();
		}
		s.pop();
	}
}
int main()
{
	memset(head, -1, sizeof(head));
	scanf("%d%d",&n,&p);
	for(int i = 1; i <= n; i++) minpay[i] = 0x7fffffff, money[i] = 0x7fffffff;
	for(int i = 1; i <= p; i++)
	{
		int u, val;
		scanf("%d%d",&u,&val);
		okbuy[u] = 1;
		money[u] = val;
	}
	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] && okbuy[i] == 1) tarjan(i);
	for(int i = 1; i <= n; i++)
		if(!dfn[i])
		{
			printf("NO
%d",i);
			return 0;
		}
	for(int i = 1; i <= n; i++)
		for(int j = head[i]; j != -1; j = e[j].next)
		{
			int v = e[j].to;
			if(color[v] != color[i])
			{
				rudu[color[v]]++;
			}
		}	
	for(int i = 1; i <= num; i++)
	{
		if(rudu[i] == 0)
		ans += minpay[i];
	}
	printf("YES
%d
",ans);
	return 0;	
}