HDU 2767 Proving Equivalences (强联通)

大意：至少加几条边让图联通。

思路：让图联通的话肯定会与度数有关。所以找到入出度最大的补上即可。

#include<map>
#include<queue>
#include<cmath>
#include<cstdio>
#include<stack>
#include<iostream>
#include<cstring>
#include<algorithm>
#define LL int
#define inf 0x3f3f3f3f
#define eps 1e-8
#include<vector>
#define ls l,mid,rt<<1
#define rs mid+1,r,rt<<1|1
using namespace std;
const int nv = 21000;
const int ne = 51000;

struct node{
    int to,next;
}q[ne<<1];

int head[ne<<1],dfn[nv],num[nv],in[nv],stk[nv],low[nv];
int cnt,top,tim,scc,out[nv],n,mi;
bool vis[nv];
void Add(int a,int b){
    q[cnt].to = b;
    q[cnt].next = head[a];
    head[a] = cnt++;
}

void Tarjan(int u){
    dfn[u] = low[u] = tim++;
    stk[top++] = u;
    vis[u] = 1;
    for(int i = head[u];~i;i=q[i].next){
        int v = q[i].to;
        if(!dfn[v]){
            Tarjan(v);
            low[u] = min(low[u],low[v]);
        }
        else if(vis[v]){
            low[u] = min(low[u],dfn[v]);
        }
    }
    if(dfn[u]== low[u]){
        scc++;
        while(top>0&&stk[top] != u){
            int now = stk[--top];
            num[now] = scc;
            vis[now] = 0 ;
        }
    }
}
void init(){
    scc =  cnt = top = 0;
    tim =  1;
    memset(head,-1,sizeof(head));
    memset(in,0,sizeof(in));
    memset(out,0,sizeof(out));
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(vis,false,sizeof(vis));
    memset(num,0,sizeof(num));
}
int main(){
    int n,m,i,j,k,cla,a,b;
    scanf("%d",&cla);
    while( cla-- ){
        init();
        scanf("%d%d",&n,&m);
        for(i = 0;i < m;++ i){
            scanf("%d%d",&a,&b);
            Add(a,b);
        }
        for(i = 1;i <= n;++ i)
            if(!dfn[i])
                Tarjan(i);
        if(scc==1){
            puts("0");continue;
        }
        for(i = 1;i <= n;++ i){
            for(j = head[i] ;~j;j = q[j].next){
                int v = q[j].to;
                if(num[i]!=num[v]){
                    out[num[i]]++;
                    in[num[v] ]++;
                }
            }
        }
        int s2 = 0,s1 = 0;
        for(i = 1;i <= scc;++ i ){
            if(!in[i])
                s1++;
            if(!out[i])
                s2++;
        }
        printf("%d\n",max(s1,s2));
    }
    return 0;
}