POJ3180（有向图强连通分量结点数>=2的个数）

The Cow Prom

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

典型的阅读理解题，看了好久的题其实就是求有向图强连通分量结点数>=2的个数。虽然简单，但是第一次确敲错了好几个小地方，敲代码时一定要专心，下面列出错误

/*
ID: LinKArftc
PROG: 3180.cpp
LANG: C++
*/

#include <map>
#include <set>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <utility>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-8
#define randin srand((unsigned int)time(NULL))
#define input freopen("input.txt","r",stdin)
#define debug(s) cout << "s = " << s << endl;
#define outstars cout << "*************" << endl;
const double PI = acos(-1.0);
const double e = exp(1.0);
const int inf = 0x3f3f3f3f;
const int INF = 0x7fffffff;
typedef long long ll;

const int maxn = 10010;
const int maxm = 50010;

struct Edge {
    int v, next;
} edge[maxm];

int head[maxn], tot;

void init() {
    tot = 0;
    memset(head, -1, sizeof(head));
}

void addedge(int u, int v) {
    edge[tot].v = v;
    edge[tot].next = head[u];
    head[u] = tot ++;
}

int n, m;
int dfn[maxn], low[maxn];
bool ins[maxn];
int scc, Time;
stack <int> st;
vector <int> vec[maxn];

void tarjan(int u) {
    int v;
    dfn[u] = low[u] = ++ Time;
    st.push(u);
    ins[u] = true;
    for (int i = head[u]; i + 1; i = edge[i].next) {
        v = edge[i].v;
        if (!dfn[v]) {
            tarjan(v);
            low[u] = min(low[v], low[u]);
        } else if (ins[v]) low[u] = min(low[u], dfn[v]);
    }
    if (dfn[u] == low[u]) {//刚开始写成dfn[u] == low[v]了
        scc ++;
        do {
            v = st.top();
            st.pop();
            ins[v] = false;
            vec[scc].push_back(v);
        } while (u != v);
    }
}

int main() {

    //input;
    int u, v;
    while (~scanf("%d %d", &n, &m)) {
        init();
        for (int i = 1; i <= m; i ++) {//刚开始写成i<=n了
            scanf("%d %d", &u, &v);
            addedge(u, v);
        }
        memset(dfn, 0, sizeof(dfn));
        memset(ins, 0, sizeof(ins));
        while (!st.empty()) st.pop();
        for (int i = 1; i <= n; i ++) vec[i].clear();
        scc = 0;
        Time = 0;
        for (int i = 1; i <= n; i ++) {
            if (!dfn[i]) tarjan(i);
        }
        int ans = 0;
        for (int i = 1; i <= scc; i ++) {
            if (vec[i].size() >= 2) ans ++;
        }
        printf("%d\n", ans);
    }

    return 0;
}