AcWing164. 可达性统计(拓扑排序+bitset)

164. 可达性统计 - AcWing题库高质量的算法题库https://www.acwing.com/problem/content/166/第一次用到bitset真是一个神奇的容器啊，让我们计算子集非常方便。

这个题也直接给出了是有个DAG所以我们可以直接求拓扑序列，然后倒序遍历将其每个点能触及到的状态记录。

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <cstring>
#include <set>
#include <cmath>
#include <map>
#include <bitset>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const int MN = 65005;
const int MAXN = 1e6 + 10;
const int INF = 0x3f3f3f3f;
#define IOS ios::sync_with_stdio(false)
#define lowbit(x) ((x)&(-x))
using P = pair<int, int>;

int ver[MAXN];
int head[MAXN];
int nxt[MAXN];
int cnt;
int in[MAXN];
int tuop[MAXN];
int cnt1;
void add(int x, int y) {
	ver[++cnt] = y;
	nxt[cnt] = head[x];
	head[x] = cnt;
	in[y]++;
}
int n, m;
const int N = 3e4 + 5;
bitset<N> rec[N];

void topsort() {
	queue<int> q;
	map<int, int> mp;
	for (int i = 1; i <= n; i++) {
		if (in[i])
			continue;
		q.push(i);
	}
	while (!q.empty()) {
		int t = q.front();
		q.pop();
		tuop[++cnt1] = t;
		for (int i = head[t]; i; i = nxt[i]) {
			int v = ver[i];
			in[v]--;
			if (!in[v])
				q.push(v);
		}
	}
}


int main() {
	scanf("%d %d", &n, &m);
	int x, y;
	for (int i = 1; i <= m; i++) {
		scanf("%d %d", &x, &y);
		add(x, y);
	}
	topsort();
	for (int i = n; i >= 1; i--) {
		int j = tuop[i];
		rec[j][j] = 1;
		for (int k = head[j]; k; k = nxt[k]) {
			int v = ver[k];
			rec[j] |= rec[v];
		}
	}
	for (int i = 1; i <= n; i++) {
		printf("%d\n", rec[i].count());
	}
	return 0;
}