浙江财经大学
信息管理与工程学院

HDU 1385 Minimum Transport Cost(Floyd)

本文由 Ocrosoft 于 2016-08-09 15:47:59 发表

Minimum Transport Cost

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10053    Accepted Submission(s): 2734

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
 

Solution
经过一个城市需要付过路费,还要按照字典序输出费用最小路径。

Floyd怎么还能这样用…理解不深刻完全不会做。推荐先理解一下Floyd->链接

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <stack>
#include <queue>
#include <ctime>
#include <string>
#include <cstdio>
#include <vector>
#include <cctype>
#include <climits>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define ms(a) memset(a,0,sizeof(a))
#define INF INT_MAX
#define MAXN 1000+10
using namespace std;
int edge[MAXN][MAXN];
int path[MAXN][MAXN];//路径
int tax[MAXN];//过路费
int main()
{
    int n;
    while(cin>>n&&n)
    {
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
            {
                path[i][j]=j;
                cin>>edge[i][j];
                if(edge[i][j]==-1)edge[i][j]=INF;
            }
        for(int i=1; i<=n; i++)cin>>tax[i];
        for(int k=1; k<=n; k++)
            for(int i=1; i<=n; i++)
                for(int j=1; j<=n; j++)
                {
                    if(edge[i][k]==INF||edge[k][j]==INF)continue;
                    int cost=edge[i][k]+edge[k][j]+tax[k];
                    if(edge[i][j]>cost)
                    {
                        edge[i][j]=cost;
                        path[i][j]=path[i][k];
                    }
                    else if(edge[i][j]==cost)
                    {
                        if(path[i][j]>path[i][k])
                            path[i][j]=path[i][k];
                    }
                }
        int s,e;
        while(cin>>s>>e&&s+1&&e+1)
        {
            printf("From %d to %d :\nPath: %d",s,e,s);
            int u=s,v=e;
            while(u!=v)
            {
                printf("-->%d",path[u][v]);
                u=path[u][v];
            }
            printf("\nTotal cost : %d\n\n",edge[s][e]);
        }
    }
    return 0;
}

 

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