LA4287 tarjan求强连通分量+缩点

题意:问至少还需要加入多少条有向边才能变成强连通图.

答案=max(in0,out0),特判scc=1.

代码:

#include<cstdio>
#include<cstring>
#include<stack>
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
using namespace std;
const int MAXE=2e5+7;
const int MAXV=2e4+7;
int V,E;
struct edge{
    int next,v;
}es[MAXE];
int head[MAXV],tot;
void init(){
    tot=0;
    memset(head,-1,sizeof(head));
}
void addEdge(int u,int v){
    es[tot].v=v;
    es[tot].next=head[u];
    head[u]=tot++;
}
int low[MAXV],dfn[MAXV],time;
int belong[MAXV],scc;
bool instack[MAXV];
stack<int>s;
void dfs(int u,int pre){
    low[u]=dfn[u]=++time;
    instack[u]=true;
    s.push(u);
    for(int i=head[u];~i;i=es[i].next){
        int v=es[i].v;
        if(!dfn[v]){
            dfs(v,u);
            low[u]=min(low[u],low[v]);
        }else if(instack[v]){
            low[u]=min(low[u],dfn[v]);
        }
    }
    if(low[u]==dfn[u]){
        scc++;
        for(;;){
            int now=s.top();s.pop();
            belong[now]=scc;
            instack[now]=false;
            if(now==u)break;
        }
    }
}
int in[MAXV],out[MAXV];
int main(){
    int T;scanf("%d",&T);
    while(T--){
        scanf("%d%d",&V,&E);    
        init();
        memset(in,0,sizeof(in));
        memset(out,0,sizeof(out));
        memset(low,0,sizeof(low));
        memset(dfn,0,sizeof(dfn));
        memset(instack,0,sizeof(instack));
        memset(belong,0,sizeof(belong));
        time=scc=0;
        for(int i=0;i<E;++i){
            int a,b;scanf("%d%d",&a,&b);
            addEdge(a,b);
        }
        for(int i=1;i<=V;++i){
            if(!dfn[i]){
                dfs(i,-1);
            }
        }
    //    printf("scc=%d
",scc);
        if(scc==1){
            printf("0
");
            continue;
        }
        for(int u=1;u<=V;++u){//缩点后的操作,对代表元进行操作
            for(int i=head[u];~i;i=es[i].next){
                int v=es[i].v;
                if(belong[u]!=belong[v]){
                    out[belong[u]]++;
                    in[belong[v]]++;
                }
            }
        }
        int IN=0,OUT=0;
        for(int i=1;i<=scc;++i){
        //    printf("in[%d]=%d,out[%d]=%d
",i,in[i],i,out[i]);
            if(in[i]==0)++IN;
            if(out[i]==0)++OUT;
        }
        printf("%d
",max(IN,OUT));
    }
    return 0;
}