hdu 1385 Minimum Transport Cost

相当于一道模板题吧。

学会了floyed 的路径保存的方法。

代码来源网络。

如下：

#include <iostream>
#include <stdio.h>
#include <memory.h>
#include <queue>
using namespace std;

const int INF = 99999999;
const int N = 505;

int map[N][N], tax[N], path[N][N];
int n;

void init()
{
    int i, j;
    for(i = 1; i <= n; i++)
        for(j = 1; j <= n; j++)
            if(i == j) map[i][j] = 0;
            else map[i][j] = INF;
}

void input()
{
    int i, j, k;
    for(i = 1; i <= n; i++)
        for(j = 1; j <= n; j++)
        {
            scanf("%d", &k);
            if(k != -1) map[i][j] = k;
            path[i][j] = j;
        }
    for(i = 1; i <= n; i++)
        scanf("%d", &tax[i]);
}

void floyd()
{
    int i, j, k, len;
    for(k = 1; k <= n; k++)
    {
        for(i = 1; i <= n; i++)
        {
            for(j = 1; j <= n; j++)
            {
                len = map[i][k] + map[k][j] + tax[k];
                if(map[i][j] > len)
                {
                    map[i][j] = len;
                    path[i][j] = path[i][k];    //标记到该点的前一个点
                }
                else if(len == map[i][j])   //若距离相同
                {
                    if(path[i][j] > path[i][k]) //判断是否为字典顺序
                        path[i][j] = path[i][k];
                }
            }
        }
    }
}

void output()
{
    int i, j, k;
    while(scanf("%d %d", &i, &j))
    {
        if(i == -1 && j == -1) break;
        printf("From %d to %d :\n", i, j);
        printf("Path: %d", i);
        k = i;
        while(k != j)   //输出路径从起点直至终点
        {
            printf("-->%d", path[k][j]);
            k = path[k][j];
        }
        printf("\n");
        printf("Total cost : %d\n\n", map[i][j]);
    }
}

int main()
{
    while(scanf("%d", &n), n)
    {
        init();
        input();
        floyd();
        output();
    }
    return 0;
}