poj 2762 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?

Input

The first line contains a single integer T, the number of test cases. And followed T cases.

The first line for each case contains two integers n, m(0 < n < 1001,m < 6000), the number of rooms and corridors in the cave. The next m lines each contains two integers u and v, indicating that there is a corridor connecting room u and room v directly.

Output

The output should contain T lines. Write ‘Yes‘ if the cave has the property stated above, or ‘No‘ otherwise.

Sample Input

1
3 3
1 2
2 3
3 1

Sample Output

Yes

Source

POJ Monthly--2006.02.26,zgl & twb

题目大意

给出一些点，和他们之间的有向边，如果图中任意两点 x，y 之间满足 x 可以到达 y 或者 y 可以到达 x ，就输出“Yes”，否则输出“No”

因为n>1000 所以挨个判断是不可取的

所以我们用tarjan 求出强连通分量的个数

然后判断这几个强连通分量在不在一条链上就可以了

tarjan缩点+拓扑排序

屠龙宝刀点击就送

#include <cstring>
#include <ctype.h>
#include <cstdio>
#include <queue>
#define M 6005
#define N 1500
using namespace std;
void read(int &x)
{
    x=0;bool f=0;
    char ch=getchar();
    while(!isdigit(ch))
    {
        if(ch==‘-‘) f=1;
        ch=getchar();
    }
    while(isdigit(ch))
    {
        x=x*10+ch-‘0‘;
        ch=getchar();
    }
    x=f?(~x)+1:x;
}
int in[N],g[N][N],tim,dfn[N],col[N],sumcol,low[N],T,n,m,head[N],cnt,stack[N],top;
bool instack[N],vis[N];
struct node
{
    int next,to;
}edge[M];
void add(int u,int v)
{
    edge[++cnt].next=head[u];
    edge[cnt].to=v;
    head[u]=cnt;
}
int min(int a,int b)
{
    return a>b?b:a;
}
void dfs(int x)
{
    stack[++top]=x;
    instack[x]=1;
    vis[x]=1;
    low[x]=dfn[x]=++tim;
    for(int i=head[x];i;i=edge[i].next)
    {
        int v=edge[i].to;
        if(instack[v]) low[x]=min(low[x],dfn[v]);
        else if(!vis[v])
        {
            dfs(v);
            low[x]=min(low[x],low[v]);
        }
    }
    if(low[x]==dfn[x])
    {
        sumcol++;
        while(x!=stack[top])
        {
            instack[stack[top]]=0;
            col[stack[top--]]=sumcol;
        }
        instack[stack[top]]=0;
        col[stack[top--]]=sumcol;
    }
}
bool judge()
{
    queue<int>q;
    for(int i=1;i<=sumcol;i++)
    if(!in[i]) q.push(i);
    if(q.size()>1) return false;
    while(!q.empty())
    {
        int now=q.front();
        q.pop();
        for(int i=1;i<=sumcol;i++)
        if(g[now][i])
        {
            in[i]--;
            if(!in[i]) q.push(i);
        }
        if(q.size()>1) return false;
    }
    return true;
}
int main()
{
    read(T);
    for(;T--;)
    {
        memset(head,0,sizeof(head));
        memset(dfn,0,sizeof(dfn));
        memset(vis,0,sizeof(vis));
        memset(instack,0,sizeof(instack));
        memset(col,0,sizeof(col));
        memset(low,0,sizeof(low));
        memset(in,0,sizeof(in));
        memset(g,0,sizeof(g));
        tim=0;top=0;sumcol=0;cnt=0;
        bool f=false;
        read(n);
        read(m);
        for(int x,y,i=1;i<=m;i++)
        {
            read(x);
            read(y);
            add(x,y);
        }
        for(int i=1;i<=n;i++)
        if(!vis[i]) dfs(i);
        if(sumcol==1) {printf("Yes\n");continue;}
        for(int i=1;i<=n;i++)
        {
            for(int j=head[i];j;j=edge[j].next)
            {
                int v=edge[j].to;
                if(i!=v&&col[i]!=col[v])
                {
                    g[col[i]][col[v]]=1;
                    in[col[v]]++;
                }
            }
        }
        if(judge()) printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}