hdu 3594 强连通判断一条边是否只属于一个环

http://acm.hdu.edu.cn/showproblem.php?pid=3594

Problem Description

1. It is a Strongly Connected graph.
2. Each edge of the graph belongs to a circle and only belongs to one circle.
We call this graph as CACTUS.



There is an example as the figure above. The left one is a cactus, but the right one isn’t. Because the edge (0, 1) in the right graph belongs to two circles as (0, 1, 3) and (0, 1, 2, 3).

 

Input

The input consists of several test cases. The first line contains an integer T (1<=T<=10), representing the number of test cases.
For each case, the first line contains a integer n (1<=n<=20000), representing the number of points.
The following lines, each line has two numbers a and b, representing a single-way edge (a->b). Each case ends with (0 0).
Notice: The total number of edges does not exceed 50000.

 

Output

For each case, output a line contains “YES” or “NO”, representing whether this graph is a cactus or not.

 

Sample Input

  

   2
4
0 1
1 2
2 0
2 3
3 2
0 0
4
0 1
1 2
2 3
3 0
1 3
0 0
  

 

Sample Output

  

   YES
NO
  
  


  

题意是说给你一个有向图，此图如果具备这两个条件：1，它是一个强连通图，2它的每一条边仅属于一个环。就输出YES,否则输出NO。
只要对tarjan算法非常了解才能够理解这么做。这个题需要很好的理解dfn和low两个数组的作用，如果low[u]!=dfn[u]，则说明已经存在环，如果我们访问u点时发现low[u]!=dfn[u]，则说明有一条边在两个环上了。

#include <cmath>
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
const int N=100005;
int stack[N],head[N],low[N],dfn[N],ins[N];
int ip,cnt_tar,top,index,ok,m,n;
struct note
{
    int to;
    int next;
};
note edge[N];
void add(int u,int v)
{
    edge[ip].to=v;edge[ip].next=head[u];head[u]=ip++;
}
void tarjan(int u)
{
    int j,i,v;
    dfn[u]=low[u]=++index;
    stack[++top]=u;
    ins[u]=1;
    for(i=head[u];i!=-1;i=edge[i].next)
    {
        v=edge[i].to;
        if(!dfn[v])
        {
            tarjan(v);
            low[u]=min(low[v],low[u]);
        }
        else if(ins[v])
        {
            low[u]=min(low[u],dfn[v]);
            if(low[v]!=dfn[v])
                   ok=1;
        }
       if(ok)return;
    }
    if(dfn[u]==low[u])
    {
        cnt_tar++;
        /*if(cnt_tar)
            ok=1;*/
        do
        {
            j=stack[top--];
            ins[j]=0;
        }
        while(j!=u);
    }
}
void solve()
{
    int i;
    top=0,index=0,cnt_tar=0;
    memset(low,0,sizeof(low));
    memset(dfn,0,sizeof(dfn));
    memset(ins,0,sizeof(ins));
    memset(stack,0,sizeof(stack));
    ok=0;
    for(i=1;i<=n;i++)
    {
        if(!dfn[i])
        {
            tarjan(i);
        }
    }
}
int main()
{
    int u,v,T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        m=0;
        memset(head,-1,sizeof(head));
        ip=0;
        while(1)
        {
            scanf("%d%d",&u,&v);
            if(u==0&&v==0)
                break;
            add(u+1,v+1);
            m++;
        }
        m-=1;
        solve();
      //  printf("ok=%d\n",ok);
       // printf("cnt_tar=%d\n",cnt_tar);
        if(ok==1||cnt_tar!=1)
        {
            printf("NO\n");
        }
        else
            printf("YES\n");
    }
    return 0;
}