CodeVS4093EZ的间谍网络

Posted 2020-07-27

tarjan缩点，找入度为0的点（可能是一条链的端点或者一个点），入度为0的点必须能被控制。

每个强连通分量中能被控制的点的最小值是缩点后这个点的最小值，入度为0的点的最小值的和就是最小花费。

还要记下每个强连通分量中的最小编号，这些编号中的最小值就是第一个不能被控制的点。

#include<iostream>
#include<cstdio>
#include<cstring>
#define N 3010
using namespace std;
struct edge
{
    int to,next;
}e[N*2],d[N*2];
int head[N],h[N];
int w[N],W[N];
bool instack[N];
int low[N],dfn[N],stack[N];
int cost[N],MIN[N];
int belong[N],rudu[N];
int Index,scc,cnt,top,r;
int n,p,x,y,u,v;
void ins(int u,int v)
{
    e[++cnt].to = v; e[cnt].next = head[u]; head[u] = cnt;
}
void tarjan(int k)
{
    dfn[k] = low[k] = ++Index;
    stack[++top] = k;
    instack[k] = true;
    for (int i=head[k];i;i = e[i].next)
    {
        int v = e[i].to;
        if (instack[v])
            low[k] = min(low[k],dfn[v]);
        else if (!dfn[v])
        {
            tarjan(v);
            low[k] = min(low[k],low[v]);
        }
    }
    if (dfn[k] == low[k])
    {
        scc++;
        int now = 0;
        while (now != k)
        {
            now = stack[top--];
            belong[now] = scc;
            instack[now] = false;
            if (w[now]) 
            {
                W[scc] = min(W[scc],w[now]);
            }
            MIN[scc] = min(MIN[scc],now);
            
        }
    }
}
int main()
{
    memset(W,127,sizeof(W));
    memset(MIN,127,sizeof(MIN));
    scanf("%d%d",&n,&p);
    for (int i=1;i<=p;i++)
    {
        scanf("%d%d",&x,&y);
        w[x] = y;
    }
    scanf("%d",&r);
    for (int i=1;i<=r;i++)
    {
        scanf("%d%d",&u,&v);
        ins(u,v);
    }
    for (int i=1;i<=n;i++)
        if (!dfn[i])
            tarjan(i);
    cnt = 0;
    for (int i=1;i<=n;i++)
    {
        for (int j = head[i];j;j = e[j].next)
        {
            int v = e[j].to;
            if (belong[i] != belong[v])
            {
                d[++cnt].to = belong[v];
                d[cnt].next = h[belong[i]];
                h[belong[i]] = ++cnt;
                rudu[belong[v]] ++;
            }
        }
    }
    int pd=0,failmin=6666666,ans = 0;
    for (int i=1;i<=scc;i++)
    {
        if (rudu[i] == 0)
        {
            if (W[i] == 2139062143)
            {
                pd = 1;
                failmin = min(failmin,MIN[i]);
            }
            ans += W[i];
        }
    }
    if (pd == 1)
    {
        printf("NO\n");
        printf("%d",failmin);
        return 0;
    }
    printf("YES\n%d",ans);
}