Floyd算法

Floyd (基于动态规划)

dist[i][j]表示从i到j最多经过k点的最小距离。
分为两种情况：经过k和不经过k:
不经过k则为dist[i][j] = dist[i][j]。
经过k则为dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j])。
由于k时逐渐增大所以dist[i][k]和dist[k][j]的最小距离之前已经得出。

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

#define x first
#define y second

using namespace std;

typedef pair<int, int> PII;
typedef long long LL;

const int N = 210, INF = 0x3f3f3f3f;

int n, m, k;
int dist[N][N];

int main() 
{
    cin >> n >> m >> k;
    memset(dist, 0x3f, sizeof dist);
    for (int i = 1; i <= n; i ++ ) dist[i][i] = 0;

    while (m -- ) 
    {
        int a, b, c;
        cin >> a >> b >> c;
        dist[a][b] = min(dist[a][b], c);
    }

    for (int k = 1; k <= n; k ++ ) 
        for (int i = 1; i <= n; i ++ ) 
            for (int j = 1; j <= n; j ++ ) 
                dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);

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

    return 0;
}