Floyd hdu1385

Minimum Transport Cost

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9546    Accepted Submission(s): 2544

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
   
   
   
   

floyd算法

1.定义概览

Floyd-Warshall算法（Floyd-Warshall algorithm）是解决任意两点间的最短路径的一种算法，可以正确处理有向图或负权的最短路径问题，同时也被用于计算有向图的传递闭包。Floyd-Warshall算法的时间复杂度为O(N3)，空间复杂度为O(N2)。

2.算法描述
1)算法思想原理：

     Floyd算法是一个经典的动态规划算法。用通俗的语言来描述的话，首先我们的目标是寻找从点i到点j的最短路径。从动态规划的角度看问题，我们需要为这个目标重新做一个诠释（这个诠释正是动态规划最富创造力的精华所在）
      从任意节点i到任意节点j的最短路径不外乎2种可能，1是直接从i到j，2是从i经过若干个节点k到j。所以，我们假设Dis(i,j)为节点u到节点v的最短路径的距离，对于每一个节点k，我们检查Dis(i,k) + Dis(k,j) < Dis(i,j)是否成立，如果成立，证明从i到k再到j的路径比i直接到j的路径短，我们便设置Dis(i,j) = Dis(i,k) + Dis(k,j)，这样一来，当我们遍历完所有节点k，Dis(i,j)中记录的便是i到j的最短路径的距离。

2).算法描述：

a.从任意一条单边路径开始。所有两点之间的距离是边的权，如果两点之间没有边相连，则权为无穷大。 　　
b.对于每一对顶点 u 和 v，看看是否存在一个顶点 w 使得从 u 到 w 再到 v 比己知的路径更短。如果是更新它。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3f
int Map[110][110],path[110][110];
int p[110];
int n;

void floyd()
{
    int dk;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            path[i][j]=j;
        }
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            for(int k=1;k<=n;k++){
                dk=Map[j][i]+Map[i][k]+p[i];
                if(dk < Map[j][k]){
                    Map[j][k]=dk;
                    path[j][k]=path[j][i];
                }
                else if(dk==Map[j][k]){
                    if(path[j][k] > path[j][i])
                        path[j][k]=path[j][i];
                }
            }
        }
    }
}
int main()
{
    int x;
    while(~scanf("%d",&n)){
        if(n==0)    break;
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                scanf("%d",&x);
                if(x==-1)   Map[i][j]=INF;
                else    Map[i][j]=x;
            }
        }
        for(int i=1;i<=n;i++)   scanf("%d",&p[i]);
        floyd();
        int sx,ed;
        while(scanf("%d%d",&sx,&ed)!=EOF){
            if(sx==-1 && ed==-1)    break;
            printf("From %d to %d :\nPath: %d",sx,ed,sx);
            x=sx;
            while(x!=ed){
                printf("-->%d",path[x][ed]);
                x=path[x][ed];
            }
            printf("\nTotal cost : %d\n\n",Map[sx][ed]);
        }
    }
    return 0;
}