acwing 854 Floyd求最短路

题面

题解

求多源最短路问题用floyd算法，O(n³), 基于动态规划，用k表示状态，每次循环都是将上一次的状态更新

代码

#include<bits/stdc++.h>

using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 210;
int n, m, k;
int d[N][N];

void init() {
    
    
    for (int i = 1; i <= n; i++) {
    
    
        for (int j = 1; j <= n; j++) {
    
    
            if (i == j) d[i][j] = 0;  //自环
            else d[i][j] = INF;   //无穷大
        }
    }
}

void floyd() {
    
    
    for (int k = 1; k <= n; k++) {
    
    
        for (int i = 1; i <= n; i++) {
    
    
            for (int j = 1; j <= n; j++) {
    
    
                d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
            }
        }
    }
}

int main() {
    
    

    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);

    cin >> n >> m >> k;

    init();

    for (int i = 1; i <= m; i++) {
    
    
        int a, b, c;
        cin >> a >> b >> c;
        d[a][b] = min(d[a][b], c);
    }

    floyd();

    while (k--) {
    
    
        int a, b;
        cin >> a >> b;
        if (d[a][b] > INF / 2) cout << "impossible" << endl;
        else cout << d[a][b] << endl;
    }

    return 0;
}