.弗洛伊德算法求最短路径

1. 5. 弗洛伊德算法求最短路径

【问题描述】

对于下面一张若干个城市，以及城市之间距离的地图，请采用弗洛伊德算法求出所有城市之间的最短路径。

【输入形式】

顶点个数n，以及n*n的邻接矩阵，其中不可达使用9999表示。

【输出形式】

每两个顶点之间的最短路径和经过的顶点

注意：规定顶点自身到自身的dist值为0，此时path为该顶点编号。

【样例输入】

4

9999 4 11 9999

6 9999 2 9999

1 9999 9999 1

9999 3 9999 9999

【样例输出】

from(空格)0(空格)to(空格)0:(空格)dist(空格)=(空格)0(空格)path:0(换行)

from 0 to 1: dist = 4 path:0 1
from 0 to 2: dist = 6 path:0 1 2
from 0 to 3: dist = 7 path:0 1 2 3
from 1 to 0: dist = 3 path:1 2 0
from 1 to 1: dist = 0 path:1
from 1 to 2: dist = 2 path:1 2
from 1 to 3: dist = 3 path:1 2 3
from 2 to 0: dist = 1 path:2 0
from 2 to 1: dist = 4 path:2 3 1
from 2 to 2: dist = 0 path:2
from 2 to 3: dist = 1 path:2 3
from 3 to 0: dist = 6 path:3 1 2 0
from 3 to 1: dist = 3 path:3 1
from 3 to 2: dist = 5 path:3 1 2
from 3 to 3: dist = 0 path:3

#include <bits/stdc++.h>
using namespace std;
const int inf = 9999;
int arc[55][55];
int n;
int dist[55][55];
vector<int> path[55][55];
void creat()
{
    cin >> n;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
        {
            cin >> arc[i][j];
            if (i == j)
                arc[i][j] = 0;
        }
}
void floyed()
{
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
        {
            dist[i][j] = arc[i][j];
            if (dist[i][j] != inf && i != j)
            {
                path[i][j].push_back(i);
                path[i][j].push_back(j);
            }
            else if (i == j)
                path[i][j].push_back(i);
            else
                path[i][j].push_back(9999);
        }
    for (int k = 0; k < n; k++)
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                if (dist[i][j] > dist[i][k] + dist[k][j])
                {
                    dist[i][j] = dist[i][k] + dist[k][j];
                    path[i][j].clear();
                    for (int q = 0; q < path[i][k].size() - 1; q++)
                    {
                        path[i][j].push_back(path[i][k][q]);
                    }
                    for (int q = 0; q < path[k][j].size(); q++)
                    {
                        path[i][j].push_back(path[k][j][q]);
                    }
                }
}
void show()
{
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
        {
            if (i == j)
            {
                dist[i][j] = 0;
                path[i][j].clear();
                path[i][j].push_back(i);
            }
            cout << "from " << i << " to " << j << ": "
                 << "dist = " << dist[i][j] << " path:";
            for (int k = 0; k < path[i][j].size(); k++)
            {
                cout << path[i][j][k] << " ";
            }
            cout << endl;
        }
}
int main()
{
    creat();
    floyed();
    show();
}