75: libreoj #10028 双向宽搜

$des$

实现一个bfs

$sol$

写了一个双向bfs

#include <bits/stdc++.h>

using namespace std;

#define Rep(i, a, b) for(int i = a; i <= b; i ++)

#define gc getchar()
inline int read() {
    int x = ; char c = gc;
    while(c < '' || c > '') c = gc;
    while(c >= '' && c <= '') x = x *  + c - '', c = gc;
    return x;
}

const int N = ;
const int xd[] = {-, -, -, -, , , , };
const int yd[] = {-, -, , , , , -, -};

int Lim = ;
int vis[N][N], bel[N][N];
int n;
int Bfs_time;
queue<pair <int, int> > Q;

inline int Work(int sx, int sy, int tx, int ty) {
    if(sx == tx && sy == ty) return ;
    memset(vis, , sizeof vis);
    memset(bel, , sizeof bel);
    while(!Q.empty()) Q.pop();
    Q.push(make_pair(sx, sy));
    Q.push(make_pair(tx, ty));
    bel[sx][sy] = , bel[tx][ty] = ;
    vis[tx][ty] = ;
    while(!Q.empty()) {
        pair <int, int> tp = Q.front();
        Q.pop();
        int px = tp.first, py = tp.second;
        Rep(i, , ) {
            int nx = px + xd[i], ny = py + yd[i];
            if(bel[nx][ny] == bel[px][py] || nx <  || nx > Lim || ny <  || ny > Lim) continue;
            if(vis[nx][ny]) return vis[nx][ny] + vis[px][py];
            vis[nx][ny] = vis[px][py] + ;
            bel[nx][ny] = bel[px][py];
            Q.push(make_pair(nx, ny));
        }
    }
}

int main() {
    n = read();
    while(n --) {
        Lim = read();
        cout << Work(read(), read(), read(), read()) << "\n";
    }
    return ;
}