首页 > ACM题库 > HDU-杭电 > HDU 1385 Minimum Transport Cost-最短路径-[解题报告] C++
2013
12-09

HDU 1385 Minimum Transport Cost-最短路径-[解题报告] C++

Minimum Transport Cost

问题描述 :

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.

输入:

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:

输出:

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.

样例输入:

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

样例输出:

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

/*
题目大意:

给你一个无向图,求任意两点的最短路。附加条件2个:

1.经过每个节点必须缴纳一点的税。

2.如果有多条最短路径,寻找字典序最小的路径。

PS:我一开始用dijstra
写了半天,WA了半天,也没搞出来个什么东东!!老是WA。。。。。。。

     后来迫不得已,改成floyd了。。。。。。。。
有种吐血的冲动~~~~

*/

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int MAXN=1202;
const int INF=99999999;
int map[MAXN][MAXN],path[MAXN][MAXN],n,cost[MAXN],st,ed;
void Floyd()
{
    int i,j,k;
    for(k=1; k<=n; k++)
        for(i=1; i<=n; i++)
            for(j=1; j<=n; j++)
            {
                if(map[i][j]>map[i][k]+map[k][j]+cost[k])
                {
                    map[i][j]=map[i][k]+map[k][j]+cost[k];
                    path[i][j]=path[i][k];
                }
                else if(map[i][j]==map[i][k]+map[k][j]+cost[k])//寻找字典序更小的路径
                {
                    if(path[i][j]>path[i][k])
                    {
                        path[i][j]=path[i][k];
                    }
                }
            }
}
int main()
{
    while(scanf("%d",&n)!=EOF&&n)
    {
        int i,j,w;
        for(i=1; i<=n; i++)
        {
            for(j=1; j<=n; j++)
            {
                path[i][j]=j;
                scanf("%d",&map[i][j]);
                if(map[i][j]==-1) map[i][j]=INF;
            }
        }
        for(i=1; i<=n; i++)
            scanf("%d",&cost[i]);
        Floyd();
        while(scanf("%d%d",&st,&ed))
        {
            if( st==-1&&ed==-1) break;
            printf("From %d to %d :\n",st,ed);
            printf("Path: %d",st);
            int u=st,v=ed;
            while (u!=v)
            {
                printf ("-->%d", path[u][v]);
                u=path[u][v];
            }
            printf("\n");
            printf("Total cost : %d\n\n",map[st][ed]);
        }
    }
    return 0;
}

解题报告转自:http://blog.csdn.net/azheng51714/article/details/8461690


  1. Good task for the group. Hold it up for every yeara??s winner. This is a excellent oppotunity for a lot more enhancement. Indeed, obtaining far better and much better is constantly the crucial. Just like my pal suggests on the truth about ab muscles, he just keeps obtaining much better.

  2. 我还有个问题想请教一下,就是感觉对于新手来说,递归理解起来有些困难,不知有没有什么好的方法或者什么好的建议?