POJ 2762 Going from u to v or from v to u? Tarjan缩点+判断链

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Going from u to v or from v to u?

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 13438 Accepted: 3507

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

13 31 22 33 1

Sample Output

Yes

Source

POJ Monthly--2006.02.26,zgl & twb

在一个图中让你判断是否任意两个点u和v，都有u能够到达v或者v能够到达u。在强联通分量里面的点肯定满足此条件，所以首先tarjan缩点，缩点之后，重新建图，要实现u到v或者v到u这一条件，那么建的新图必须满足是一条链，如果不是链，而在u点有分支，一边连接v1，一边连接v2的话，那么v1不可能到达v2，v2也不可能到达v1，因此就不能满足条件，所以不成立。

//4616K547MS#include<stdio.h>#include<string.h>#include<algorithm>#include<queue>#define M 1007using namespace std;int dfn[M],low[M],head[M],vis[M],stack[M],belong[M];int n,m,cnt,scnt,begin,num;int g[M][M],in[M];struct E{    int v,to;}edg[M*M];void init(){    memset(head,-1,sizeof(head));    memset(dfn,0,sizeof(dfn));    memset(low,0,sizeof(low));    memset(belong ,0,sizeof(belong));    memset(stack,0,sizeof(stack));    memset(vis,0,sizeof(vis));    memset(g,0,sizeof(g));    memset(in,0,sizeof(in));    cnt=scnt=num=begin=0;}void addedge(int u,int v){    edg[num].v=v;edg[num].to=head[u];    head[u]=num++;}void tarjan(int x){    int v;    dfn[x]=low[x]=++cnt;    stack[++begin]=x;    for(int i=head[x];i!=-1;i=edg[i].to)    {        v=edg[i].v;        if(!dfn[v])        {            tarjan(v);            low[x]=min(low[x],low[v]);        }        else if(!vis[v])            low[x]=min(low[x],dfn[v]);    }    if(low[x]==dfn[x])    {        scnt++;        do        {            v=stack[begin--];            belong[v]=scnt;            vis[v]=1;        }while(v!=x);    }}bool link()//判断能否形成一条链{    queue<int>q;    for(int i=1;i<=scnt;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<=scnt;i++)            if(g[now][i])            {                in[i]--;                if(!in[i])q.push(i);            }        if(q.size()>1)return false;    }    return true;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        init();        scanf("%d%d",&n,&m);        int a,b;        for(int i=1;i<=m;i++)        {            scanf("%d%d",&a,&b);            addedge(a,b);        }        for(int i=1;i<=n;i++)            if(!dfn[i])tarjan(i);        if(scnt==1) {printf("Yes\n");continue;}        for(int u=1;u<=n;u++)//缩点之后，重新建图            for(int j=head[u];j!=-1;j=edg[j].to)            {                int v=edg[j].v;                if(u!=v&&belong[u]!=belong[v])                {                    g[belong[u]][belong[v]]=1;                    in[belong[v]]++;                }            }        if(link())printf("Yes\n");        else printf("No\n");    }    return 0;}