可达性统计
原题链接:可达性统计
题目大意
给你一张\(n\)个点\(m\)条边的有向无环图,分别统计从每个点出发能够到达的点的数量
题目题解
看到题意就知道要用到拓扑排序,但是拓扑排序的理论复杂度在30000的极限条件下会超时,这个时候我们考虑使用 \(bitset\),一个很好用的代替bool的防卡常技巧,详细的说明这里不说,可以去百度上查看相关运用
//#define fre yes
#include <queue>
#include <bitset>
#include <cstdio>
#include <cstring>
#include <algorithm>
const int N = 30005;
int head[N << 1], to[N << 1], ver[N << 1];
int in[N], ans[N];
std::bitset<N> f[N];
int tot, k;
void addedge(int x, int y) {
ver[tot] = y;
to[tot] = head[x];
head[x] = tot++;
}
int n, m;
void toposort() {
std::queue<int> q;
for (int i = 1; i <= n; i++) {
if(!in[i]) q.push(i);
}
while(!q.empty()) {
int t = q.front(); q.pop();
ans[++k] = t;
for (int i = head[t]; ~i; i = to[i]) {
int v = ver[i];
in[v]--;
if(!in[v]) q.push(v);
}
}
}
int main() {
memset(head, -1, sizeof(head));
scanf("%d %d", &n, &m);
for (int i = 1; i <= m; i++) {
int u, v;
scanf("%d %d", &u, &v);
addedge(u, v);
in[v] ++;
} toposort();
// for (int i = 1; i <= n; i++) {
// printf("%d ", ans[i]);
// }
for (int i = n; i >= 1; i--) {
int j = ans[i];
f[j][j] = 1;
for (int x = head[j]; ~x; x = to[x]) {
f[j] |= f[ver[x]];
}
}
for (int i = 1; i <= n; i++) {
printf("%d\n", f[i].count());
} return 0;
}
原文地址:https://www.cnblogs.com/Nicoppa/p/11549865.html
时间: 2024-11-01 11:38:49