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

看来自己floyd还是没有深刻的理解....用Floyd存路径很是不错...get....

 #include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <string>
#include <cstring>
#include <queue>
using namespace std;
#define INF 0x3f3f3f3f
#define MAX 111 int cost[MAX];
int mp[MAX][MAX],path[MAX][MAX];
int n; void floyd()
{
for(int k=; k<=n; k++)
for(int i=; i<=n; i++)
for(int j=; j<=n; j++){
int ans=mp[i][k]+mp[k][j]+cost[k];
if(mp[i][j]>ans){
mp[i][j]=ans;
path[i][j]=path[i][k];
}
else if(mp[i][j]==ans && path[i][j]>path[i][k]){
path[i][j]=path[i][k];
}
}
} int main()
{
while(~scanf("%d",&n)&&n!=){
for(int i=; i<=n; i++){
for(int j=; j<=n; j++){
if(i!=j)
mp[i][j]=INF;
else
mp[i][j]=;
path[i][j]=j;
}
}
for(int i=; i<=n; i++){
for(int j=; j<=n; j++){
scanf("%d",&mp[i][j]);
if(mp[i][j]==-)
mp[i][j]=INF;
}
}
for(int i=; i<=n; i++){
scanf("%d",&cost[i]);
}
floyd();
int a,b;
while(~scanf("%d%d",&a,&b)){
if(a==-&&b==-)
break;
printf("From %d to %d :\n",a,b);
printf("Path: ");
int s=a;
while(s!=b){
printf("%d-->",s);
s=path[s][b];
}
printf("%d\n",b);
printf("Total cost : %d\n\n", mp[a][b]);
} }
}

NSOJ Minimum Transport Cost的更多相关文章

  1. HDU 1385 Minimum Transport Cost (Dijstra 最短路)

    Minimum Transport Cost http://acm.hdu.edu.cn/showproblem.php?pid=1385 Problem Description These are ...

  2. Minimum Transport Cost(floyd+二维数组记录路径)

    Minimum Transport Cost Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/O ...

  3. HDU1385 Minimum Transport Cost (Floyd)

    Minimum Transport Cost Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/O ...

  4. hdu 1385 Minimum Transport Cost(floyd &amp;&amp; 记录路径)

    Minimum Transport Cost Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/O ...

  5. hdu 1385 Minimum Transport Cost (Floyd)

    Minimum Transport CostTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Ot ...

  6. ZOJ 1456 Minimum Transport Cost(Floyd算法求解最短路径并输出最小字典序路径)

    题目链接: https://vjudge.net/problem/ZOJ-1456 These are N cities in Spring country. Between each pair of ...

  7. Minimum Transport Cost Floyd 输出最短路

    These are N cities in Spring country. Between each pair of cities there may be one transportation tr ...

  8. hdu 1385 Minimum Transport Cost (floyd算法)

    貌似···················· 这个算法深的东西还是很不熟悉!继续学习!!!! ++++++++++++++++++++++++++++ ======================== ...

  9. HDU 1385 Minimum Transport Cost (最短路,并输出路径)

    题意:给你n个城市,一些城市之间会有一些道路,有边权.并且每个城市都会有一些费用. 然后你一些起点和终点,问你从起点到终点最少需要多少路途. 除了起点和终点,最短路的图中的每个城市的费用都要加上. 思 ...

随机推荐

  1. Cocos2dx 3.0开发环境的搭建--Eclipse建立在Android工程

    一.前言: 这部分描述了Cocos2d-x 3.0的一些基础内容,以及在Eclipse上上编译我们的Cocos2d-x项目,成功把Helloworld执行起来了.看完本篇博客之后.你就会知道Cocos ...

  2. 我理解设计模式C++实现观察者模式Observer Pattern

    概述: 近期中国股市起起伏伏,当然了起伏就用商机,小明发现商机后果断想入市,买入了中国证券,他想在电脑client上,网页上,手机上,iPad上都能够查看到该证券的实时行情,这样的情况下我们应该怎么设 ...

  3. clientIDMode的应用

    以前我们可以通过ClientID在JavaScript脚本中服务器端控件.比如: document.getElementById("<%=控件.ClientID %>" ...

  4. 乐在其中设计模式(C#) - 提供者模式(Provider Pattern)

    原文:乐在其中设计模式(C#) - 提供者模式(Provider Pattern) [索引页][源码下载] 乐在其中设计模式(C#) - 提供者模式(Provider Pattern) 作者:weba ...

  5. 乐在其中设计模式(C#) - 组合模式(Composite Pattern)

    原文:乐在其中设计模式(C#) - 组合模式(Composite Pattern) [索引页][源码下载] 乐在其中设计模式(C#) - 组合模式(Composite Pattern) 作者:weba ...

  6. VC档(夹)文件夹路径的经营方针和代码

    ***********************************************声明*************************************************** ...

  7. Directx11学习笔记【五】 基本的数学知识----向量篇

    本文参考dx11龙书 Chapter1 vector algebra(向量代数) 要想学好游戏编程,扎实的数学知识是尤为重要的,下面将对dx11龙书中有关向量的数学知识做一下总结. 在数学中,几何向量 ...

  8. UVA 538 - Balancing Bank Accounts(贪心)

    UVA 538 - Balancing Bank Accounts 题目链接 题意:给定一些人的欠钱关系,要求在n-1次内还清钱,问方案 思路:贪心,处理出每一个人最后钱的状态,然后直接每一个人都和最 ...

  9. MySQL InnoDB数据库备份与还原

    备份 进入cm黑窗口 输入下列命令 mysqldump -u 用户名 -p 数据库名称> c:\11.sql 回车执行 恢复 进入cm黑窗口 输入下列命令 mysql>use dbtest ...

  10. Java泛型Restletclient

    写一个与restletserver通信的client类.用于測试通信是否成功.而且进行交互.为了方便其它人使用.于是,写一个通用的方法封装起来,但是中途却放生了一些问题. 依照正常写法,顺序走下来是这 ...