854. Floyd求最短路 - AcWing题库

floyd算法用来求多源汇最短路问题。

初始的时候d数组是邻接矩阵，d[i][j]存放所有的边。

然后用一个三层for循环：

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

循环完之后d[i][j]存的就是i到j的最短路的长度。

code

#define _CRT_SECURE_NO_WARNINGS 1. 
#include<bits/stdc++.h>

using namespace std;

const int N = 210, INF = 1e9;

int n, m, Q;
int d[N][N];

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()
{
    cin >> n >> m >> Q;

    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            if (i == j) d[i][j] = 0;
            else d[i][j] = INF;

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

    floyd();

    while (Q--)
    {
        int a, b;
        cin >> a >> b;

        int t = d[a][b];
        if (t > INF / 2) puts("impossible");
        else printf("%d\n", t);
    }

    return 0;
}