题面

题解

可以发现, 若 \(u\) 到 \(v\) 可由从 \(u\) 到 \(v\) 的另外一条路径而不走 \((u, v)\) 边到达, 则 \((u, v)\) 可删

正反用 bitset 维护一下传递闭包即可

Code

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <bitset>
#include <queue>
const int N = 30005; 
const int M = 1e5 + 5; 
using namespace std; 

int n, m, cnte[2], head[2][N], in[2][N], ans; 
struct edge { int to, nxt; } e[2][M]; 
bitset<N> bit[2][N]; 
queue<int> q; 

template < typename T >
inline T read()
{
    T x = 0, w = 1; char c = getchar(); 
    while(c < '0' || c > '9') { if(c == '-') w = -1; c = getchar(); }
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); 
    return x * w; 
}

inline void adde(int u, int v, int opt) { e[opt][++cnte[opt]] = (edge) { v, head[opt][u] }, head[opt][u] = cnte[opt]; } 

void toposort(int opt)
{
    for(int i = 1; i <= n; i++)
        if(!in[opt][i]) q.push(i); 
    while(!q.empty())
    {
        int u = q.front(); q.pop(); 
        for(int v, i = head[opt][u]; i; i = e[opt][i].nxt)
        {
            v = e[opt][i].to; bit[opt][v] |= bit[opt][u], bit[opt][v][u] = 1; 
            in[opt][v]--; if(!in[opt][v]) q.push(v); 
        }
    }
}

int main()
{
    n = read <int> (), m = read <int> (); 
    for(int u, v, i = 1; i <= m; i++)
    {
        u = read <int> (), v = read <int> (); 
        in[0][v]++, in[1][u]++; 
        adde(u, v, 0), adde(v, u, 1); 
    }
    toposort(0), toposort(1); 
    for(int i = 1; i <= m; i++)
        if((bit[0][e[0][i].to] & bit[1][e[1][i].to]).any()) ans++; 
    printf("%d\n", ans); 
    return 0; 
}