POJ 2762 Going from u to v or from v to u?(Tarjan)

题意：对任给的两个点x，y，是否一定有路，从x到y，或者从y到x，，，刚开始看错了，以为是赤裸裸的Tarjan，结果是或，不是且，，比较简单

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
using namespace std;
const int N = 1009;
const int M = 6009;
const int INF = 0x3f3f3f3f;
int n,m;
struct LT{
    int nex,to;
} L[M];
int F[N],cnt;
void add(int f,int t)
{
    L[cnt].nex = F[f];
    L[cnt].to = t;
    F[f] =cnt++;
}
void init()
{
    scanf("%d%d",&n,&m);
    memset(F,0,sizeof(F));
    cnt =1;
    int f,t;
    for(int i=0;i<m;i++)
    scanf("%d%d",&f,&t),add(f,t);
}
int dfn[N],low[N],post[N],col[N],color,ind;
stack<int> S;
void tarjan(int k)
{
    dfn[k] = low[k] = ind++;
    post[k] = 1;S.push(k);
    for(int i=F[k];i;i=L[i].nex)
    {
        int to = L[i].to;
        if(!dfn[to])
        {
            tarjan(to);
            low[k] = min(low[k],low[to]);
        }else if(post[to]&&dfn[to]<low[k])
        {
            low[k] = dfn[to];
        }
    }
    if(low[k]==dfn[k])
    {
        int i;color++;
        for(i=S.top(),S.pop();i!=k;i=S.top(),S.pop())
        {
            post[i] = 0,col[i] = color;
        }
        post[k] = 0;col[k] = color;
    }
}
struct node{
    int nex,to;
} tL[M];
int tF[N],tcnt;
int in[N];
void tadd(int f,int t)
{
    tL[tcnt].nex = tF[f];
    tL[tcnt].to = t;
    tF[f] = tcnt++;
}
void solve()
{
    memset(dfn,0,sizeof(dfn));
    color = 0;
    ind = 1;
    for(int i=1;i<=n;i++)
    if(!dfn[i]) tarjan(i);
    memset(tF,0,sizeof(tF));
    memset(in,0,sizeof(in));
    tcnt =1;
    for(int i=1;i<=n;i++)
    for(int j=F[i];j;j=L[j].nex)
    {
        int to = L[j].to;
        if(col[i]!=col[to])
        {
            tadd(col[i],col[to]);
        }
    }
    memset(in,0,sizeof(in));
    for(int i=1;i<tcnt;i++) in[tL[i].to]++;

    int ans =0,v=-1;
    for(int j=1;j<color;j++)
    {//for(int i=1;i<=color;i++) cout<<in[i]<<" ";cout<<endl;
        ans =0;
        for(int i=1;i<=color;i++)
        if(!in[i])
        {
            ans++;in[i] = INF;v=i;
        }
        if(ans>1)
        {
            printf("No\n");return ;
        }
        for(int i=tF[v];i;i=tL[i].nex)
        {
            in[tL[i].to]--;
        }
    }
    printf("Yes\n");
}
int main()
{
    freopen("in.txt","r",stdin);
    int cas;
    scanf("%d",&cas);
    while(cas--)
    {
        init();
        solve();

    }
    return 0;
}