JDOJ-1076: Popular Cow

1076: Popular Cow

Time Limit: 1 Sec  Memory Limit: 13 MB
Submit: 386  Solved: 131
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Description

　　Every cow has a dream: to be the most popular famous cow in a group! In a herd with N (1<=N<=10,000) cattle, give you M (1<=M<=50,000) tuples (A,B), indicating that A thinks B is popular . Since popularity is transitive, if A thinks B is popular and B thinks C is popular, then A also thinks C is popular, even if it's not a very clear rule. Your task is to count the number of cows that are liked by all other cows.

Input

　　The first line, two numbers, N and M. Lines 2 to M+1, each line has two numbers, A and B, indicating that A thinks B is popular.

Output

　　A number, the number of cows considered popular by all other cows.

Sample Input

3 3 1 2 2 1 2 3

Sample Output

1

HINT

 

[Example description] Cow #3 is the only one considered famous by all other cows.

 

Summary: Tarjan shrinks the points, if the number of points with an out-degree of 0 is 0, ans = n, if the number of points with an out-degree of 0 is 1, ans = sum[this point], if the number is greater than 1, ans = 0.

#include<bits/stdc++.h>

using namespace std;
const int maxn = 100005;

int dfn[maxn], low[maxn], head[maxn << 1];
bool vis[maxn]; int  n, m, cnt = 1, sum[maxn];

struct Node{
    int v, nxt;
}G[maxn << 1];

void insert(int u, int v) {
    G[cnt] = (Node) {v, head[u]}; head[u] = cnt++;
}

stack<int> s;
int tot = 0, color[maxn], tim = 0;
void tarjan(int x) {
    vis[x] = true; s.push(x);
    dfn[x] = low[x] = ++tim;
    for (int i = head[x]; i; i = G[i].nxt) {
        int v = G[i].v;
        if(!dfn[v]) {
            tarjan(v); low[x] = min(low[x], low[v]); 
        }
        else if(vis[v]) low[x] = min(low[x], dfn[v]); 
    }
    if(dfn[x] == low[x]) {
        tot++;
        while(1) {
            int now = s.top();
            vis[now] = false; s.pop();
            color[now] = tot;
            sum[tot]++;
            if(now == x) break;
        }
    }
}
int chu[maxn];

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= m; ++i) {
        int x, y;
        scanf("%d%d", &x, &y); insert(x, y);
    }
    for (int i = 1; i <= n; ++i) if(!dfn[i]) tarjan(i);
    for (int u = 1; u <= n; ++u) {
        for (int i = head[u]; i; i = G[i].nxt) {
            int v = G[i].v;
            if(color[u] != color[v]) chu[color[u]]++;
        }
    } int ans = 0; int fl = 0;
    for (int i = 1; i <= tot; ++i) if(chu[i] == 0) ans = sum[i], fl++;
    if(fl > 1) printf("0\n"); else if(fl == 0) printf("%d\n", n);
    else printf("%d\n", ans);
    return 0;
}