poj3180 The Cow Prom

The Cow Prom

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2373		Accepted: 1402

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

题目大意：找点数≥2的强连通分量.

分析：套模板.

#include <cstdio>
#include <stack>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = ;

int n,m,head[maxn],to[maxn],nextt[maxn],tot = ,pre[maxn],low[maxn],scc[maxn],cnt,du[maxn],dfs_clock;
int ans,numm,num[maxn];
stack <int> s;

void add(int x,int y)
{
    to[tot] = y;
    nextt[tot] = head[x];
    head[x] = tot++;
}

void tarjan(int u)
{
    s.push(u);
    pre[u] = low[u] = ++dfs_clock;
    for (int i = head[u];i;i = nextt[i])
    {
        int v = to[i];
        if (!pre[v])
        {
            tarjan(v);
            low[u] = min(low[u],low[v]);
        }
        else
            if (!scc[v])
                low[u] = min(low[u],pre[v]);
    }
    if (pre[u] == low[u])
    {
        cnt++;
        while()
        {
            int t = s.top();
            s.pop();
            scc[t] = cnt;
            num[cnt]++;
            if(t == u)
                break;
        }
    }
}

int main()
{
    scanf("%d%d",&n,&m);
    for(int i = ; i <= m; i++)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        add(a,b);
    }
    for (int i = ; i <= n; i++)
        if (!pre[i])
            tarjan(i);
    for (int i = ; i <= cnt; i++)
        if (num[i] >= )
        ans++;
    printf("%d\n",ans);

    return ;
}