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

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

   题目大意：给你一个有向图，让你判断任意两个顶点 u和v 之间是否存在可达的路径，即 u可以到达v 或 v 可以到达 u 。

   解题思路：先用tarjan缩点，然后用拓扑排序一下，排序后的序列中，如果所有相邻两个点之间均存在边，则说明原图中，任意两点之间均存在可达的路径，即输出“Yes”，否则，输出“No”。

   请看代码：

#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std ;
const int MAXN = 1111 ;
struct Node
{
    int adj ;
    Node *next ;
}mem[MAXN * 6]; // 先把存放边节点的数组开好，这样能大量节省时间
int memp ;
Node *vert[MAXN] ; // 建立顶点数组
short g[MAXN][MAXN] ; // 缩点后的邻接矩阵
bool inq[MAXN] ; // 判断节点是否在栈中
int stap[MAXN] ; // 数组模拟栈
int top ;
int d[MAXN] ;  // 统计 缩点后各连通分量的 入度
int tpo[MAXN] ;  // 拓扑序列
bool vis[MAXN] ;
int dfn[MAXN] ;
int low[MAXN] ;
int tmpdfn ;
int belong[MAXN] ;
int n , m ;
int scnt ;  // 记录强连通分量个数
void clr() // 初始化
{
    memset(dfn , 0 , sizeof(dfn)) ;
    memset(low , 0 , sizeof(low)) ;
    memset(vert , 0 , sizeof(vert)) ;
    memset(belong , -1 , sizeof(belong)) ;
    memset(vis , 0 , sizeof(vis)) ;
    memset(g , 0 , sizeof(g)) ;
    memset(stap , -1 , sizeof(stap)) ;
    memset(d , 0 , sizeof(d)) ;
    memset(tpo , -1 , sizeof(tpo)) ;
    memset(inq , 0 , sizeof(inq)) ;
    memp = 0 ;
    top = -1 ;
    scnt = -1 ;
    tmpdfn = 0 ;
}
void tarjan(int u)
{
    vis[u] = 1 ;
    dfn[u] = low[u] = ++ tmpdfn ;
    stap[++ top] = u ;
    inq[u] = true ;
    Node *p = vert[u] ;
    while (p != NULL)
    {
        int v = p -> adj ;
        if(!vis[v])
        {
            tarjan(v) ;
            low[u] = min(low[u] , low[v]) ;
        }
        else if(inq[v])  // 如果点v还在栈中
        {
            low[u] = min(low[u] , dfn[v]) ;
        }
        p = p -> next ;
    }
    if(dfn[u] == low[u])  // 缩点
    {
        scnt ++ ;
        int tmp = stap[top --] ;
        do
        {
            tmp = stap[top --] ;
            inq[tmp] = false ;
            belong[tmp] = scnt ;
        }while(tmp != u) ;
    }
}
void build()
{
    Node * p ;
    int i ;
    for(i = 1 ; i <= n ; i ++)
    {
        p = vert[i] ;
        while (p)
        {
            int tv = p -> adj ;
            int x = belong[i] ;
            int y = belong[tv] ;
            g[belong[i]][belong[tv]] = 1 ;
            if(x != y)
                d[y] ++ ;
            p = p -> next ;
        }
    }
}
void topo()  // 拓扑排序
{
    int k ;
    for(k = 0 ; k <= scnt ; k ++)
    {
        int i ;
        for(i = 0 ; i <= scnt ; i ++)
        {
            if(d[i] == 0)
                break ;
        }
        tpo[k] = i ;
        d[i] = -1 ;
        int j ;
        for(j = 0 ; j <= scnt ; j ++)
        {
            if(i != j)
                d[j] -= g[i][j] ;
        }
    }
}
void init()
{
    scanf("%d%d" , &n , &m) ;
    int i , j ;
    clr() ;
    int root ;
    for(i = 0 ; i < m ; i ++)
    {
        int a , b ;
        scanf("%d%d" , &a , &b) ;
        Node *p = &mem[memp] ;
        p -> adj = b ;
        p -> next = vert[a] ;
        vert[a] = p ;

        memp ++ ;
    }
    for(i = 1 ; i <= n ; i ++)  // 注意此处，保证图中每个节点都被访问
    {
        if(!vis[i])
        tarjan(i) ;
    }
    build() ;
    topo() ;
    bool flag = true ;
    for(i = 0 ; i < scnt ; i ++)
    {
        if(!g[ tpo[i] ][ tpo[i + 1] ])
        {
            flag = false ;
            break ;
        }
    }
    if(flag)
        puts("Yes") ;
    else
        puts("No") ;
}
int main()
{
    int T ;
    scanf("%d" , &T) ;
    while (T --)
    {
        init() ;
    }
    return 0 ;
}