间谍网络

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终于有一道能够一次AC的Tarjan题啦！

这个题还是很简单的，首先Tarjan缩点，之后把新图建出来之后发现，因为要控制所有间谍，那么肯定得从入度为0的间谍下手，所以如果有任何入度为0的间谍不愿意被收买的话任务就失败了，否则的话直接topo排序统计答案即可。

图也许是不联通的，所以要用循环的方法一直跑即可。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<queue>
#include<set>
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define enter putchar('\n')

using namespace std;
typedef long long ll;
const int M = 50005;

int read()
{
    int ans = 0,op = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
    if(ch == '-') op = -1;
    ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
    ans *= 10;
    ans += ch - '0';
    ch = getchar();
    }
    return ans * op;
}

struct edge
{
    int next,to,from;
}e[M],e1[M];

int n,p,r,a,b,low[M],dfn[M],stack[M],scc[M],mmoney[M],mnode[M],head[M],ecnt,idx,cnt,top;
int rdeg[M],cdeg[M],ecnt1,head1[M],mo[M],ans;
bool vis[M],pd[M],can[M],jud[M];

void add(int x,int y)
{
    e[++ecnt].to = y;
    e[ecnt].from = x;
    e[ecnt].next = head[x];
    head[x] = ecnt;
}

void add1(int x,int y)
{
    e[++ecnt1].to = y;
    e[ecnt1].from = x;
    e[ecnt1].next = head1[x];
    head1[x] = ecnt1;
}

void tarjan(int x)
{
    low[x] = dfn[x] = ++idx;
    stack[++top] = x,vis[x] = 1;
    for(int i = head[x];i;i = e[i].next)
    {
    if(!dfn[e[i].to]) tarjan(e[i].to),low[x] = min(low[x],low[e[i].to]);
    else if(vis[e[i].to]) low[x] = min(low[x],dfn[e[i].to]);
    }
    if(low[x] == dfn[x])
    {
    cnt++;
    int p;
    while((p = stack[top--]))
    {
        scc[p] = cnt,vis[p] = 0;
        if(can[p]) jud[cnt] = 1,mmoney[cnt] = (!mmoney[cnt]) ? mo[p] : min(mmoney[cnt],mo[p]);
        mnode[cnt] = (!mnode[cnt]) ? p : min(p,mnode[cnt]);
        if(p == x) break;
    }
    }
}

void rebuild()
{
    rep(i,1,ecnt)
    {
    int r1 = scc[e[i].from],r2 = scc[e[i].to];
    //printf("@%d %d\n",r1,r2);
    if(r1 != r2) add1(r1,r2),rdeg[r2]++,cdeg[r1]++;
    }
}

void topo(int x)
{
    if(!jud[x])
    {
    printf("NO\n");
    printf("%d\n",mnode[x]);
    exit(0);
    }
    queue <int> q;
    q.push(x);pd[x] = 1,ans += mmoney[x];
    while(!q.empty())
    {
    int k = q.front();q.pop();
    for(int i = head1[k];i;i = e1[i].next)
    {
        cdeg[k]--,rdeg[e1[i].to]--;
        if(!rdeg[e1[i].to]) q.push(e1[i].to);
    }
    }
}

int main()
{
    n = read(),p = read();
    rep(i,1,p) a = read(),b = read(),can[a] = 1,mo[a] = b;
    r = read();
    rep(i,1,r) a = read(),b = read(),add(a,b);
    //rep(i,1,ecnt) printf("!%d %d\n",e[i].from,e[i].to);
    rep(i,1,n) if(!dfn[i]) tarjan(i);
    rebuild();
    //rep(i,1,ecnt1) printf("#%d %d\n",e1[i].from,e1[i].to);
    rep(i,1,cnt) if(!pd[i] && !rdeg[i]) topo(i);
    printf("YES\n");
    printf("%d\n",ans);
    return 0;
}