acwing-164. 可达性统计

考点

  • 拓扑排序
  • 状态压缩

题解

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#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e4 + 50;
int n, m, cnt, tot = -1, ind[maxn], head[maxn], seq[maxn];
queue<int> q;
bitset<maxn> ans[maxn];

struct node {
int v_, nxt_;
} e[2 * maxn];

void add(int u, int v) {
e[++tot].v_ = v, e[tot].nxt_ = head[u];
head[u] = tot;
}

void toposort() {
for (int i = 1; i <= n; ++i) {
if (ind[i] == 0) q.push(i);
}
while (!q.empty()) {
int u = q.front();
q.pop();
// 这里不要使用1 << u,最多只有32位,而这里是3e4位
ans[u].set(u);
seq[++cnt] = u;
for (int i = head[u]; ~i; i = e[i].nxt_) {
int v = e[i].v_;
if (--ind[v] == 0) q.push(v);
}
}
}

int main() {
cin >> n >> m;
int u, v;
memset(head, -1, sizeof(head));
for (int i = 1; i <= m; ++i) cin >> u >> v, add(u, v), ++ind[v];
toposort();
for (int i = cnt; i; --i) {
int u = seq[i];
for (int j = head[u]; ~j; j = e[j].nxt_) {
int v = e[j].v_;
ans[u] |= ans[v];
}
}
for (int i = 1; i <= n; ++i) cout << ans[i].count() << endl;
return 0;
}

思路

直接上图。