BZOJ 1051 受欢迎的牛

tarjan缩点

由题意可以发现，其实最受欢迎的牛就是找出所有强联通分量，缩点之后，所有新点里出度为0的那个强联通分量里的所有点。。。
而且这中强联通分量只能有一个，假设大于1个的话，就会有两块不联通的区域，那就无法被所有牛喜欢。。
所以直接缩点计算一下出度就行啦

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 10005;
int n, m, cnt, head[N], dfn[N], low[N], k, ins[N], tot, scc[N], d[N];
struct Edge { int v, next; } edge[N<<5];
stack<int> st;
vector<int> c[N];
 
void addEdge(int a, int b){
    edge[cnt].v = b, edge[cnt].next = head[a], head[a] = cnt ++;
}
 
void tarjan(int s){
    dfn[s] = low[s] = ++k;
    ins[s] = true, st.push(s);
    for(int i = head[s]; i != -1; i = edge[i].next){
        int u = edge[i].v;
        if(!dfn[u]){
            tarjan(u);
            low[s] = min(low[s], low[u]);
        }
        else if(ins[u])
            low[s] = min(low[s], dfn[u]);
    }
    if(dfn[s] == low[s]){
        tot ++; int cur = 0;
        do{
            cur = st.top(); st.pop();
            ins[cur] = false, scc[cur] = tot, c[tot].push_back(cur);
        }while(cur != s);
    }
}
 
int main(){
 
    full(head, -1);
    n = read(), m = read();
    for(int i = 0; i < m; i ++){
        int a = read(), b = read();
        addEdge(a, b);
    }
    for(int i = 1; i <= n; i ++){
        if(!dfn[i]) tarjan(i);
    }
    for(int s = 1; s <= n; s ++){
        for(int i = head[s]; i != -1; i = edge[i].next){
            int u = edge[i].v;
            if(scc[s] != scc[u]) d[scc[s]] ++;
        }
    }
    int ans = 0;
    for(int i = 1; i <= tot; i ++){
        if(!d[i]){
            if(ans) { printf("0\n"); return 0;}
            ans = i;
        }
    }
    //printf(num != 1 ? "0" : "%d\n", ans);
    cout << c[ans].size() << endl;
    return 0;
}