# 强连通分量

$tarjan将强连通分量缩点转化成拓扑图的原理看$这里

$最受欢迎的牛必然是拓补图的终点,且若图中不止一个终点,必然没有某个牛被所有牛欢迎$

#include <bits/stdc++.h>
using namespace std;
#define IO ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
inline int lowbit(int x) { return x & (-x); }
#define ll long long
#define pb push_back
#define PII pair<int, int>
#define x first
#define y second
#define inf 0x3f3f3f3f
const int N = 10010, M = 50010;
int n, m;
int h[N], e[M], ne[M], idx;
int dfn[N], low[N], timestamp;
int stk[N], top;
bool in_stk[N];
int id[N], scc_cnt, Size[N];
int dout[N];

void add(int a, int b) {
e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

void tarjan(int u) {
dfn[u] = low[u] = ++timestamp;
stk[++top] = u, in_stk[u] = true;
for (int i = h[u]; ~i; i = ne[i]) {
int j = e[i];
if (!dfn[j]) {
tarjan(j);
low[u] = min(low[u], low[j]);
}
else if (in_stk[j]) low[u] = min(low[u], dfn[j]);
}
if (dfn[u] == low[u]) {
++scc_cnt;
int y;
do {
y = stk[top--];
in_stk[y] = false;
id[y] = scc_cnt;
Size[scc_cnt]++;
} while (y != u);
}
}

int main() {
IO;
cin >> n >> m;
memset(h, -1, sizeof h);
while (m--) {
int a, b;
cin >> a >> b;
}
for (int i = 1; i <= n; ++i)
if (!dfn[i]) tarjan(i);

for (int i = 1; i <= n; ++i)
for (int j = h[i]; ~j; j = ne[j]) {
int k = e[j];
int a = id[i], b = id[k];
if (a != b) dout[a]++;
}

int zeros = 0, sum = 0;
for (int i = 1; i <= scc_cnt; ++i)
if (!dout[i]) {
zeros++;
sum += Size[i];
if (zeros > 1) {
sum = 0;
break;
}
}
cout << sum << '\n';
return 0;
}

posted @ 2021-02-20 01:42  phr2000  阅读(4)  评论(0编辑  收藏