CH2101 可达性统计(算竞进阶习题)

拓扑排序+状态压缩

考虑每一个点能够到达的所有点都是与该店相邻的点的后继节点，可知：

令f[u]表示u点可到达的节点个数，f[u]={u}与f[v](u, v)的并集

于是可以利用状态压缩，能够到达的节点用1表示，这样更新f的时候直接求并集即可

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C yql){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % yql)if(p & 1)ans = 1LL * x * ans % yql;
    return ans;
}
const int N = 30005;
int n, m, cnt, head[N], ind[N], tps[N], tot;
bitset<N> f[N];
struct Edge{
    int v, next;
}edge[N<<2];

void addEdge(int a, int b){
    edge[cnt].v = b;
    edge[cnt].next = head[a];
    head[a] = cnt ++;
}

void bfs(){
    queue<int> q;
    for(int i = 1; i <= n; i ++){
        if(ind[i] == 0) q.push(i);
    }
    while(!q.empty()){
        int s = q.front(); q.pop();
        tps[tot++] = s;
        for(int i = head[s]; i != -1; i = edge[i].next){
            int u = edge[i].v;
            if(--ind[u] == 0) q.push(u);
        }
    }
}

int main(){

    memset(head, -1, sizeof head);
    n = read(), m = read();
    for(int i = 0; i < m; i ++){
        int a = read(), b = read();
        addEdge(a, b), ind[b] ++;
    }
    bfs();
    for(int i = tot - 1; i >= 0; i --){
        int u = tps[i];
        f[u][u] = 1;
        for(int j = head[u]; j != -1; j = edge[j].next){
            int v = edge[j].v;
            f[u] |= f[v];
        }
    }
    for(int i = 1; i <= n; i ++)
        printf("%d
", f[i].count());
    return 0;
}