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

Floyd判斷是否一步到達

將一步到達的連變跑Floyd

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

using namespace std;
const int N = 70;
const int oo = 999999999;

bool f[N][N][N];
int n, m, now = 1, head[N], dis[N];
 
struct Node{int u, v, w, nxt;} G[N * N];
int Que[N * 5];
bool vis[N];
int U[N][N];

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

inline int read() {
    int x = 0, f = 1; char c = yxy;
    while(c < ‘0‘ || c > ‘9‘) {if(c == ‘-‘) f = -1; c = yxy;}
    while(c >= ‘0‘
 
 && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = yxy;
    return x * f;
}

int main() {
    n = read();
    m = read();
    memset(U, 120, sizeof U);
    for(int i = 1; i <= m; i ++) {
        int u = read(), v = read();
        f[u][v][0] = 1;
        U[u][v] = 1;
    }
    for(int k = 1
 
; k <= 36; k ++)
        for(int i = 1; i <= n; i ++)
            for(int j = 1; j <= n; j ++)
                for(int t = 1; t <= n; t ++) {
                    if((f[i][t][k - 1] == 1) && (f[t][j][k - 1] == 1)) {
                        f[i][j][k] = 1;
                        U[i][j] = 1;
                    }
                }
    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= n; j ++)
            for(int k = 1; k <= n; k ++)
                if(U[i][k] + U[k][j] >= 0) U[i][j] = min(U[i][j], U[i][k] + U[k][j]);
    cout << U[1][n];
    return 0;
}

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