首页 > ACM题库 > HDU-杭电 > HDU 1690 Bus System-最短路径-[解题报告] C++
2013
12-21

HDU 1690 Bus System-最短路径-[解题报告] C++

Bus System

问题描述 :

Because of the huge population of China, public transportation is very important. Bus is an important transportation method in traditional public transportation system. And it’s still playing an important role even now.
The bus system of City X is quite strange. Unlike other city’s system, the cost of ticket is calculated based on the distance between the two stations. Here is a list which describes the relationship between the distance and the cost.

Your neighbor is a person who is a really miser. He asked you to help him to calculate the minimum cost between the two stations he listed. Can you solve this problem for him?
To simplify this problem, you can assume that all the stations are located on a straight line. We use x-coordinates to describe the stations’ positions.

输入:

The input consists of several test cases. There is a single number above all, the number of cases. There are no more than 20 cases.
Each case contains eight integers on the first line, which are L1, L2, L3, L4, C1, C2, C3, C4, each number is non-negative and not larger than 1,000,000,000. You can also assume that L1<=L2<=L3<=L4.
Two integers, n and m, are given next, representing the number of the stations and questions. Each of the next n lines contains one integer, representing the x-coordinate of the ith station. Each of the next m lines contains two integers, representing the start point and the destination.
In all of the questions, the start point will be different from the destination.
For each case,2<=N<=100,0<=M<=500, each x-coordinate is between -1,000,000,000 and 1,000,000,000, and no two x-coordinates will have the same value.

输出:

For each question, if the two stations are attainable, print the minimum cost between them. Otherwise, print “Station X and station Y are not attainable.” Use the format in the sample.

样例输入:

2
1 2 3 4 1 3 5 7
4 2
1
2
3
4
1 4
4 1
1 2 3 4 1 3 5 7
4 1
1
2
3
10
1 4

样例输出:

Case 1:
The minimum cost between station 1 and station 4 is 3.
The minimum cost between station 4 and station 1 is 3.
Case 2:
Station 1 and station 4 are not attainable.

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1690

分析:求出任意两点这间的最小消费.对m次询问就可直接打出来.

#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<iomanip>

using namespace std;
const int maxn=1000+10;
const __int64 inf=100000000002;

__int64 L1,L2,L3,L4,C1,C2,C3,C4;
__int64 map[maxn][maxn];
__int64 d[maxn];

int main(){
    int T; cin>>T;
    int cas=1;
    while(T--){
        cin>>L1>>L2>>L3>>L4>>C1>>C2>>C3>>C4;
        int n,m; cin>>n>>m;
        ///初始化
        for(int i=1;i<=n;++i){
            cin>>d[i];
            map[i][i]=0;
            for(int j=1;j<i;++j){
                __int64 s=max(d[i],d[j])-min(d[i],d[j]),v;
                if(s>L4) v=inf;
                else if(s>L3) v=C4;
                else if(s>L2) v=C3;
                else if(s>L1) v=C2;
                else if(s>0)  v=C1;
                else v=0;
                map[i][j]=map[j][i]=v;
            }
        }
        ///floyd算法
        for(int k=1;k<=n;++k)
            for(int i=1;i<=n;++i)
                for(int j=1;j<=n;++j)
                    map[i][j]=min(map[i][j],map[i][k]+map[k][j]);

        printf("Case %d:\n",cas++);
        while(m--){
            int x,y; cin>>x>>y;
            if(map[x][y]==inf)printf("Station %d and station %d are not attainable.\n",x,y);
            else printf("The minimum cost between station %d and station %d is %I64d.\n",x,y,map[x][y]);
        }
    }
    return 0;
}

解题报告转自:http://blog.csdn.net/du489380262/article/details/8914049


  1. 这道题目虽然简单,但是小编做的很到位,应该会给很多人启发吧!对于面试当中不给开辟额外空间的问题不是绝对的,实际上至少是允许少数变量存在的。之前遇到相似的问题也是恍然大悟,今天看到小编这篇文章相见恨晚。