hdu 1385 Minimum Transport Cost 最短路 + 打印字典序最小路径

Minimum Transport Cost

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Problem Description

These are N cities in Spring country. Between each pair of cities there may be one transportation track or none. Now there is some cargo that should be delivered from one city to another. The transportation fee consists of two parts:

The cost of the transportation on the path between these cities, and

a certain tax which will be charged whenever any cargo passing through one city, except for the source and the destination cities.

You must write a program to find the route which has the minimum cost.



Input

First is N, number of cities. N = 0 indicates the end of input.

The data of path cost, city tax, source and destination cities are given in the input, which is of the form:

a11 a12 ... a1N

a21 a22 ... a2N

...............

aN1 aN2 ... aNN

b1 b2 ... bN

c d

e f

...

g h

where aij is the transport cost from city i to city j, aij = -1 indicates there is no direct path between city i and city j. bi represents the tax of passing through city i. And the cargo is to be delivered from city c to city d, city e to city f, ..., and
g = h = -1. You must output the sequence of cities passed by and the total cost which is of the form:



Output

From c to d :

Path: c-->c1-->......-->ck-->d

Total cost : ......

......

From e to f :

Path: e-->e1-->..........-->ek-->f

Total cost : ......

Note: if there are more minimal paths, output the lexically smallest one. Print a blank line after each test case.



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

题意：

有N个城市，然后直接给出这些城市之间的邻接矩阵，矩阵中-1代表那两个城市无道路相连，其他值代表路径长度。

如果一辆汽车经过某个城市，必须要交一定的钱(可能是过路费)。

现在要从a城到b城，花费为路径长度之和，再加上除起点与终点外所有城市的过路费之和。

求最小花费，如果有多条路经符合，则输出字典序最小的路径。

这是一道很好的最短路题目。
题目的关键是按照字典序输出字典序最小路径。

假设有

1--->2 2

2--->3 1

1--->3 3

求1到3的最小花费路径.

那么显然后两条路:

1-->3 3

1-->2-->3 3

它们的花费是相同的，但是路径的字典序不同，“123”比“13”字典序要小，所以应当输出1-->2-->3.
用path[i][j]用来保存 i --> j 的最短路径中 i 的最优后驱（即最近），在Floyd三重循环时，一直更新path。

由于刚开始学最短路，打印路径时参考了别人的代码。

#include<stdio.h>
#include<string.h>
#define INF 1<<25
const int N = 1000;
int path

, d

, tax
;
//path[i][j]用来保存 i --> j 的最短路径中 i 的最优后驱（即最近），在Floyd三重循环时，一直更新path。
void Floyd(int n)
{
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            path[i][j] = j;
    for(int k = 1; k <= n; k++)
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
            {
                if(d[i][k] < INF && d[k][j] < INF)
                {
                    int temp = d[i][k] + d[k][j] + tax[k];
                    if(temp < d[i][j])
                    {
                        d[i][j] = temp;
                        path[i][j] = path[i][k];
                    }
                    else if(temp == d[i][j])
                    {
                        if(path[i][j] > path[i][k])
                            path[i][j] = path[i][k];
                    }
                }

            }
}

int main()
{
    int n, i, j, s, t;
    while(~scanf("%d",&n) && n)
    {
        for(i = 1; i <= n; i++)
            for(j = 1; j <= n; j++)
            {
                 scanf("%d",&d[i][j]);
                 if(d[i][j] == -1)
                    d[i][j] = INF;
            }
        for(i = 1; i <= n; i++)
            scanf("%d",&tax[i]);
        Floyd(n);
        while(~scanf("%d%d",&s,&t))
        {
            if(s == -1 && t == -1)   break;
            printf("From %d to %d :\n", s, t);
            printf("Path: %d",s);
            int temp = s;
            while(temp != t)
            {
                printf("-->%d",path[temp][t]);
                temp = path[temp][t];
            }
            printf("\n");
            printf("Total cost : %d\n",d[s][t]);
            printf("\n");
        }
    }
    return 0;
}