POJ 3180 强联通缩点

The Cow Prom

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1010		Accepted: 633

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

 

求强连通分量内部节点数>=2的连通分量个数，大水题一发。

代码：

 

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#include<stack>
using namespace std;
const int maxn=50010;
int head[10*maxn],tol,indexx,top,scc,ans,low[maxn],dfn[maxn],belong[maxn],Stack[maxn],instack[maxn];
struct node
{
        int next,to;
}edge[600000];
void add(int u,int v)
{
        edge[tol].to=v;
        edge[tol].next=head[u];
        head[u]=tol++;
}
void tarjin(int u)
{
        low[u]=dfn[u]=++indexx;
        instack[u]=1;Stack[top++]=u;
        int i,v;
        for(i=head[u];i!=-1;i=edge[i].next)
        {
                v=edge[i].to;
                if(!dfn[v])
                {
                        tarjin(v);
                        if(low[u]>low[v])low[u]=low[v];
                }
                else if(instack[v]&&low[u]>dfn[v])low[u]=dfn[v];
        }
        if(low[u]==dfn[u])
        {
                scc++;
                int cnt=0;
                do
                {
                        v=Stack[--top];
                        instack[v]=0;
                        cnt++;
                }while(u!=v);
                if(cnt>=2)ans++;
        }
}
int main()
{
        int i,j,k,m,n;
        while(~scanf("%d%d",&n,&m))
        {
                memset(head,-1,sizeof(head));tol=0;
                while(m--)
                {
                        scanf("%d%d",&i,&j);
                        add(i,j);
                }
                memset(dfn,0,sizeof(dfn));
                memset(instack,0,sizeof(instack));
                indexx=top=scc=0;
                ans=0;
                for(i=1;i<=n;i++)if(!dfn[i])tarjin(i);
                printf("%d\n",ans);
        }
        return 0;
}