# 洛谷 [P2341] 受欢迎的牛

## 强连通分量

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <stack>
using namespace std;
const int MAXN = 100005;
int init() {
int rv = 0, fh = 1;
char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') fh = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
rv = (rv<<1) + (rv<<3) + c - '0';
c = getchar();
}
return fh * rv;
}
int n, m, nume, head[MAXN], dfn[MAXN], low[MAXN], ind, cnt, fa[MAXN], num[MAXN], out[MAXN], ans;
bool f[MAXN];
struct edge{
int to, nxt;
}e[MAXN];
void adde(int from, int to) {
e[++nume].to = to;
}
stack <int> sta;
void tarjan(int u) {
low[u] = dfn[u] = ++ind;
f[u] = 1;
sta.push(u);
for(int i = head[u]; i; i = e[i].nxt) {
int v = e[i].to;
if(!dfn[v]) {
tarjan(v);
low[u] = min(low[u], low[v]);
}else if(f[v]) low[u] = min(low[u], dfn[v]);
}
if(dfn[u] == low[u]) {
int v;
cnt++;
do {
v = sta.top();
fa[v] = cnt;
f[v] = 0;
num[cnt]++;
sta.pop();
} while(v != u);
}
}
int main() {
n = init(); m = init();
for(int i = 1; i <= m; i++) {
int u = init(), v = init();
}
for(int i = 1; i <= n; i++) {
if(!dfn[i]) tarjan(i);
}
for(int i = 1; i <= n; i++) {
for(int j = head[i]; j; j = e[j].nxt) {
int v = e[j].to;
if(fa[i] != fa[v]) out[fa[i]]++;
}
}
for(int i = 1; i <= cnt; i++) {
if(!out[i]) {
if(ans) {ans = 0;break;}
ans = num[i];
}
}
cout << ans << endl;
return 0;
}

posted @ 2018-05-24 16:23  Mr_Wolfram  阅读(74)  评论(0编辑  收藏