【floyd记录路径】HDU 1385 Minimum Transport Cost

http://acm.hdu.edu.cn/showproblem.php?pid=1385

题意: 找一个路径使得从一个地方坐的士到另一个地方花费最小，其中除了起点和终点，途中经过的站点要收费

Sample Input
5
0 3 22 -1 4
3 0 5 -1 -1
22 5 0 9 20
-1 -1 9 0 4
4 -1 20 4 0
5 17 8 3 1
1 3
3 5
2 4
-1 -1
0

Sample Output
From 1 to 3 :
Path: 1-->5-->4-->3
Total cost : 21

From 3 to 5 :
Path: 3-->4-->5
Total cost : 16

From 2 to 4 :
Path: 2-->1-->5-->4
Total cost : 17

#include <iostream>
#include <fstream>
#include <algorithm>
#include <string>
#include <set>
//#include <map>
#include <queue>
#include <utility>
#include <stack>
#include <list>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
//#include <ctime>
#include <ctype.h>
using namespace std;
#define inf 999999999    //用0x3fffffff太大溢出，纠结了良久
#define M 205

int dist[M][M], n, b[M], path[M][M];

void floyd ()
{
    int i, j, k;
    for (i = 1; i <= n; i++)
        for (j = 1; j <= n; j++)
            path[i][j] = j;    //记录路径数组初始化，表示从i到j经过的第一个站
    for (k = 1; k <= n; k++)
    {
        for (i = 1; i <= n; i++)
        {
            for (j = 1; j <= n; j++)
            {
                int tp = dist[i][k] + dist[k][j] + b[k];
               //溢出之地，因为+了b[k]，其实先判断是否有路会更保险
                if (tp < dist[i][j])
                    dist[i][j] = tp, path[i][j] = path[i][k];
                else if (tp == dist[i][j] && path[i][k] < path[i][j])
                    path[i][j] = path[i][k];    //按字典序记录路径

            }
        }
    }
}

int main()
{
    int i, j, u, v, tp;
    while (scanf ("%d", &n), n)
    {
        for (i = 1; i <= n; i++)
        {
            for (j = 1; j <= n; j++)
            {
                scanf ("%d", dist[i]+j);
                if (dist[i][j] == -1)
                    dist[i][j] = inf;
            }
        }
        for (i = 1; i <= n; i++)
            scanf ("%d", b+i);
        floyd();
        while (scanf ("%d%d", &u, &v), (u != -1 || v != -1))
        {
            printf ("From %d to %d :\n", u, v);
            printf ("Path: %d", u);
            tp = u;
            while (u != v)
            {
                printf ("-->%d", path[u][v]);
                u = path[u][v];
            }
            printf ("\nTotal cost : %d\n\n", dist[tp][v]);
        }
    }
    return 0;
}