POJ 3180 The Cow Prom （求强连通分量结点数>=2的个数）

The Cow Prom

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 643		Accepted: 384

Description

The N (2 <= N <= 10,000) cows are so excited: it's prom night! They are dressed in their finest gowns, complete with corsages and new shoes. They know that tonight they will each try to perform the Round Dance.

Only cows can perform the Round Dance which requires a set of ropes and a circular stock tank. To begin, the cows line up around a circular stock tank and number themselves in clockwise order consecutively from 1..N. Each cow faces the tank so she can see the other dancers.

They then acquire a total of M (2 <= M <= 50,000) ropes all of which are distributed to the cows who hold them in their hooves. Each cow hopes to be given one or more ropes to hold in both her left and right hooves; some cows might be disappointed.

For the Round Dance to succeed for any given cow (say, Bessie), the ropes that she holds must be configured just right. To know if Bessie's dance is successful, one must examine the set of cows holding the other ends of her ropes (if she has any), along with the cows holding the other ends of any ropes they hold, etc. When Bessie dances clockwise around the tank, she must instantly pull all the other cows in her group around clockwise, too. Likewise,
if she dances the other way, she must instantly pull the entire group counterclockwise (anti-clockwise in British English).

Of course, if the ropes are not properly distributed then a set of cows might not form a proper dance group and thus can not succeed at the Round Dance. One way this happens is when only one rope connects two cows. One cow could pull the other in one direction, but could not pull the other direction (since pushing ropes is well-known to be fruitless). Note that the cows must Dance in lock-step: a dangling cow (perhaps with just one rope) that is eventually pulled along disqualifies a group from properly performing the Round Dance since she is not immediately pulled into lockstep with the rest.

Given the ropes and their distribution to cows, how many groups of cows can properly perform the Round Dance? Note that a set of ropes and cows might wrap many times around the stock tank.

Input

Line 1: Two space-separated integers: N and M

Lines 2..M+1: Each line contains two space-separated integers A and B that describe a rope from cow A to cow B in the clockwise direction.

Output

Line 1: A single line with a single integer that is the number of groups successfully dancing the Round Dance.

Sample Input

5 4
2 4
3 5
1 2
4 1

Sample Output

1

Hint

Explanation of the sample:

ASCII art for Round Dancing is challenging. Nevertheless, here is a representation of the cows around the stock tank:

       _1___

      /**** \

   5 /****** 2

  / /**TANK**|

  \ \********/

   \ \******/  3

    \ 4____/  /

     \_______/

Cows 1, 2, and 4 are properly connected and form a complete Round Dance group. Cows 3 and 5 don't have the second rope they'd need to be able to pull both ways, thus they can not properly perform the Round Dance.

Source

USACO 2006 January Silver

 

题目看了好久没有看懂，然后看了大牛的解题报告，说是求结点大于等于2的强连通分量个数

 

于是套用模板：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
#define MAXN 10005 //结点的最大数
struct Node
{
    int to,next;
}edge1[MAXN*5],edge2[MAXN*5];
//edge1用来存原图G，edge2用来存GT，即逆图
//都是用邻接表来储存的图，
int head1[MAXN];//原图的邻接表的头结点
int head2[MAXN];//逆图的邻接表的头结点
int mark1[MAXN],mark2[MAXN];
int tot1,tot2;
int cnt1,cnt2,st[MAXN],belong[MAXN];
int num,setNum[MAXN];
struct Edge
{
    int l,r;
}e[MAXN*5];//储存边 

void add(int a,int b)//添加边a->b，在原图和逆图都需要添加
{
    edge1[tot1].to=b;edge1[tot1].next=head1[a];head1[a]=tot1++;//邻接表存的原图
    edge2[tot2].to=a;edge2[tot2].next=head2[b];head2[b]=tot2++;//邻接表存的逆图 
}   
void DFS1(int x)//深度搜索原图 
{
    mark1[x]=1;
    for(int i=head1[x];i!=-1;i=edge1[i].next)
          if(!mark1[edge1[i].to])  DFS1(edge1[i].to);
    st[cnt1++]=x;//st数组是按照完成时间从小到大排序的 
}      
void DFS2(int x)//深度搜索逆图
{
    mark2[x]=1;
    num++;
    belong[x]=cnt2;
    for(int i=head2[x];i!=-1;i=edge2[i].next)
       if(!mark2[edge2[i].to]) DFS2(edge2[i].to);
}     
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        tot1=tot2=1;
        for(int i=1;i<=n;i++)//初始化 
        {
            head1[i]=head2[i]=-1;
            mark1[i]=mark2[i]=0;
        }
        for(int i=1;i<=m;i++)
        {
            int w,v;
            scanf("%d%d",&w,&v);
            e[i].l=w;e[i].r=v;//储存边 
            add(w,v);//建立邻接表 
        }        
        cnt1=cnt2=1;
        for(int i=1;i<=n;i++)
        {
            if(!mark1[i])DFS1(i);
        }    
        for(int i=cnt1-1;i>=1;i--)
        {
            if(!mark2[st[i]])
            {
                num=0;
                DFS2(st[i]);
                setNum[cnt2++]=num;
            }    
        }
        
        int cnt=0;
        for(int i=1;i<cnt2;i++)
            if(setNum[i]>=2)cnt++;
        printf("%d\n",cnt);     
    }    
    return 0;
}