POJ 2762 判断无向图的弱连通

Going from u to v or from v to u?

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 16859 Accepted: 4502

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

意：给出n个山洞，对于每个山洞，如果任意选择两点s,e，都满足s可以到达e或者e可以到达s，则输出Yes,否则输出No。

分析：实际上判断是否弱连通，所以首先强连通，然后缩点，对缩点形成的图最多只能有一个入度为0的点，如果有多个入度为

0的点，则这两个连通分量肯定是不连通的。缩点后形成的图形是一棵树，入度为0的点是这颗树的根，这棵树只能是单链，不

能有分叉，如果有分叉，则这些分叉之间是不可达的，所以就对这棵树进行DFS，如果是单链则是YES。

#include <iostream>#include <string.h>#include <stdio.h>using namespace std;const int N = 10005;struct Edge{    int to;    Edge *next;};Edge *map1[N],*map2[N];  //分别保存原图和缩点后的图int dfn[N],low[N],stack[N],belong[N],indeg[N];int index,scc_num,top;bool tmp[N];int result[N];void Tarjan(int u){    dfn[u] = low[u] = ++index;    stack[++top] = u;    tmp[u] = true;    for(Edge *p = map1[u]; p; p = p->next) //枚举每一条边    {        int v = p->to;        if(!dfn[v])        {            Tarjan(v);  //dfs继续下找            low[u] = min(low[u],low[v]);        }        else if(tmp[v])        {            low[u] = min(low[u],dfn[v]);        }    }    if(low[u] == dfn[u]) //如果节点u是强连通分量的根    {        int t;        ++scc_num;  //强连通分量个数加1        do        {            t = stack[top--];            tmp[t] = false;            belong[t] = scc_num;  //记录属于第几个强连通分量        }        while(t != u);    }}int Count(int n){    for(int i=1; i<=n; i++)        if(!dfn[i])            Tarjan(i);    return scc_num;}int Find() //在新图中找入度为0的点,如果只有一个就返回位置，否则返回0{    int record;    int cnt = 0;    for(int i=1; i<=scc_num; i++)    {        if(indeg[i] == 0)        {            cnt++;            record = i;        }    }    if(cnt == 1) return record;    return 0;}bool TopSort(){    int u,num = 0;    while(u = Find())    {        result[num++] = u;        indeg[u] = -1;        Edge *p = map2[u];        while(p)        {            indeg[p->to]--;            p = p->next;        }    }    if(num == scc_num) return true;    return false;}void Init(){    index = 0;    top = 0;    scc_num = 0;    memset(dfn,0,sizeof(dfn));    memset(low,0,sizeof(low));    memset(indeg,0,sizeof(indeg));    memset(tmp,false,sizeof(tmp));    memset(map1,NULL,sizeof(map1));    memset(map2,NULL,sizeof(map2));    memset(result,0,sizeof(result));}int main(){    int T,m,n;    scanf("%d",&T);    while(T--)    {        Init();        scanf("%d%d",&n,&m);        while(m--)        {            int a,b;            scanf("%d%d",&a,&b);            Edge *p = new Edge();            p->to = b;            p->next = map1[a];            map1[a] = p;        }        int cnt = Count(n);        if(cnt == 1)        {            puts("Yes");            continue;        }        for(int i=1; i<=n; i++)        {            Edge *p = map1[i];            while(p)            {                if(belong[i] != belong[p->to])                {                    indeg[belong[p->to]]++;                    Edge *q = new Edge();                    q->to = belong[p->to];                    q->next = map2[belong[i]];                    map2[belong[i]] = q;                }                p = p->next;            }        }        bool flag = false;        int ans = 0;        for(int i=1; i<=cnt; i++)        {            if(indeg[i] == 0)                ans++;        }        if(ans > 1) flag = false;        else if(TopSort()) flag = true;        if(flag) puts("Yes");        else     puts("No");    }    return 0;}