【缩点+拓扑判链】POJ2762 Going from u to v or from v to u?

Description

In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors connecting some rooms. Each time, Wind choose two rooms x and y, and ask one of their little sons go from one to the other. The son can either go from x to y, or from y to x. Wind promised that her tasks are all possible, but she actually doesn't know how to decide if a task is possible. To make her life easier, Jiajia decided to choose a cave in which every pair of rooms is a possible task. Given a cave, can you tell Jiajia whether Wind can randomly choose two rooms without worrying about anything?

 

Solution

缩点，满足题意的充要条件是存在一条完整的链，这个可以用“是否存在唯一拓扑序”解决。

Code

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=1e5+5;

int pre[maxn],low[maxn],clock;
int scc[maxn],s[maxn],c,cnt;
int head[maxn],e[maxn],nxt[maxn],k;
void adde(int u,int v){
    e[++k]=v;nxt[k]=head[u];head[u]=k;
}
int _head[maxn],_e[maxn],_nxt[maxn],_k;
void _adde(int u,int v){
    _e[++_k]=v;_nxt[_k]=_head[u];_head[u]=_k;
}
int n,m;

int r[maxn],vis[maxn];
int topo(){
    for(int i=1;i<=cnt;i++){
        int tot=0,u;
        for(int i=1;i<=cnt;i++)//偷懒
            if(!r[i]&&!vis[i]) tot++,u=i;
        if(tot>1) return 0;
        vis[u]=1;
        for(int i=_head[u];i;i=_nxt[i])
            r[_e[i]]--;
    }
    return 1;
}

void dfs(int u){
    pre[u]=low[u]=++clock;
    s[++c]=u;
    for(int i=head[u];i;i=nxt[i]){
        int v=e[i];
        if(!pre[v]){
            dfs(v);
            low[u]=min(low[u],low[v]);
        }
        else if(!scc[v]){
            low[u]=min(low[u],pre[v]);
        }
    }
    if(low[u]==pre[u]){
        cnt++;
        while(c){
            scc[s[c]]=cnt;
            if(s[c--]==u) break;
        }
    }
}

void clear(){
    memset(head,0,sizeof(head));
    memset(_head,0,sizeof(_head));
    memset(pre,0,sizeof(pre));
    memset(low,0,sizeof(low));
    memset(vis,0,sizeof(vis));
    memset(r,0,sizeof(r));
    c=cnt=k=_k=0;
}

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        clear();
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;i++){
            int u,v;
            scanf("%d%d",&u,&v);
            adde(u,v);
        }
        for(int i=1;i<=n;i++)
            if(!pre[i]) dfs(i);
            
        for(int i=1;i<=n;i++)
            for(int j=head[i];j;j=nxt[j]){
                int u=scc[i],v=scc[e[j]];
                if(u==v) continue;
                else _adde(u,v),r[v]++;
            }
        
        if(topo()) printf("Yes
");
        else printf("No
");
    }
    return 0;
}