hdu1385Minimum Transport Cost(最短路变种)

题目链接：

huangjing

思路：

输出路径的最短路变种问题。。这个题目在于多组询问。那么个人认为用floyd更加稳妥一点。还有就是在每一个城市都有过路费，所以在floyd的时候更改一下松弛条件就可以。。那么输出路径怎么办呢？？我採用的是输出起点的后继而不是终点的前驱。。由于我们关心的是路径字典序最小，关心的是起点的后继。。。那么打印路径的时候就直接从前向后打印，这个和dijkstra的打印路径稍有不同。。。

最短路的打印參见传送门

题目：

Minimum Transport Cost

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7538    Accepted Submission(s): 1935

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

 

Source

Asia 1996, Shanghai (Mainland China)

 

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代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#define INF 0x3f3f3f3f
using namespace std;

const int maxn=50+10;
int dis[maxn][maxn],path[maxn][maxn],n,cost[maxn];
int u,st,en;

void floyd()
{
    for(int k=1;k<=n;k++)
      for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
    {
        int tmp=dis[i][k]+dis[k][j]+cost[k];
        if(tmp<dis[i][j]||(tmp==dis[i][j]&&path[i][j]>path[i][k]))
        {
            dis[i][j]=tmp;
            path[i][j]=path[i][k];
        }
    }
}

void read_Graph()
{
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
    {
        scanf("%d",&u);
        if(u==-1)
            dis[i][j]=INF;
        else
        {
            dis[i][j]=u;
            path[i][j]=j;
        }
    }
    for(int i=1;i<=n;i++)
       scanf("%d",&cost[i]);
}

void solve()
{
    while(~scanf("%d%d",&st,&en))
    {
        if(st==-1&&en==-1)  break;
        printf("From %d to %d :
",st,en);
        printf("Path: %d",st);
        int Gery=st;
        while(Gery!=en)
        {
            printf("-->%d",path[Gery][en]);
            Gery=path[Gery][en];
        }
        printf("
Total cost : %d

",dis[st][en]);
    }
}

int main()
{
    while(~scanf("%d",&n),n)
    {
        read_Graph();
        floyd();
        solve();
    }
    return 0;
}