最短路

Minimum Transport Cost

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 21   Accepted Submission(s) : 7

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

50 3 22 -1 43 0 5 -1 -122 5 0 9 20-1 -1 9 0 44 -1 20 4 05 17 8 3 11 33 52 4-1 -10

 

Sample Output

From 1 to 3 :Path: 1-->5-->4-->3Total cost : 21From 3 to 5 :Path: 3-->4-->5Total cost : 16From 2 to 4 :Path: 2-->1-->5-->4Total cost : 17

 

题意：给你一个n*n的矩阵，s[i][j]代表从i到j的花费如果s[i][j]==-1，代表i到j不可达。每个城市的花费cost[i].求从st到ed的最小花费，利用Floyed(),注意要按照字典序的顺序输出。#include<stdio.h>#include<string.h>#define maxint 0xfffffff#define max 1000int path[max][max];int vis[max];int cost[max];int s[max][max];int st,ed,t;void Floyed(){    int i,j,k;    int temp;    for(k=1; k<=t; k++)        for(i=1; i<=t; i++)            for(j=1; j<=t; j++)            {                temp=s[i][k]+s[k][j]+cost[k];                //标记该点的前一个点，若距离相等，判断是否为字典顺序                if(temp<s[i][j]||(temp==s[i][j]&&path[i][k]<path[i][j]))                {                   s[i][j]=temp;                   path[i][j]=path[i][k];                }            }}int main(){    int temp;    while(scanf("%d",&t),t)    {        for(int i=1; i<=t; i++)            for(int j=1; j<=t; j++)            {                scanf("%d",&temp);                s[i][j]=temp;                if(temp==-1)                    s[i][j]=maxint;                path[i][j]=j;            }        for(int i=1; i<=t; i++)            scanf("%d",&cost[i]);        Floyed();        while(scanf("%d%d",&st,&ed)!=EOF)        {            if(st==ed&&st==-1)                break;            printf("From %d to %d :\n",st,ed);            printf("Path: %d",st);            int gh=st;            while(st!=ed)            {                printf("-->%d",path[st][ed]);               st=path[st][ed];            }            printf("\n");            printf("Total cost : %d\n",s[gh][ed]);            printf("\n");        }    }    return 0;}