[Luogu] 跑路

https://www.luogu.org/problemnew/show/P1613

Floyd判断是否一步到达

将一步到达的连变跑Floyd

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;
const int N = ;
const int oo = ;

bool f[N][N][N];
int n, m, now = , head[N], dis[N];
struct Node{int u, v, w, nxt;} G[N * N];
int Que[N * ];
bool vis[N];
int U[N][N];

#define yxy getchar()
#define RR freopen("gg.in", "r", stdin)

inline int read() {
    int x = , f = ; char c = yxy;
    while(c < '' || c > '') {if(c == '-') f = -; c = yxy;}
    while(c >= '' && c <= '') x = x *  + c - '', c = yxy;
    return x * f;
}

int main() {
    n = read();
    m = read();
    memset(U, , sizeof U);
    for(int i = ; i <= m; i ++) {
        int u = read(), v = read();
        f[u][v][] = ;
        U[u][v] = ;
    }
    for(int k = ; k <= ; k ++)
        for(int i = ; i <= n; i ++)
            for(int j = ; j <= n; j ++)
                for(int t = ; t <= n; t ++) {
                    if((f[i][t][k - ] == ) && (f[t][j][k - ] == )) {
                        f[i][j][k] = ;
                        U[i][j] = ;
                    }
                }
    for(int i = ; i <= n; i ++)
        for(int j = ; j <= n; j ++)
            for(int k = ; k <= n; k ++)
                if(U[i][k] + U[k][j] >= ) U[i][j] = min(U[i][j], U[i][k] + U[k][j]);
    cout << U[][n];
    return ;
}