浙江财经大学
信工学院ACM集训队

HDU 1690 Bus System

本文由 Ocrosoft 于 2016-08-11 22:48:27 发表

Bus System

Problem Description

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.

Input

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.

Output

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.

Sample Input

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

Sample Output

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.

Solution

别看题目很长,其实是水题。
题意:有n个车站,每个车站离你所在的位置为dist,票价根据dist依据上表改变。输出询问的两个车站的最少花费。输入第一行是数据组数,接下来是l1,l2,l3…c3,c4,接下来n行是dist,再下去m行是询问的两个车站。
思路:因为车站最多100个,所以可以求出两个车站的距离,算出价格,构建邻接矩阵,然后就是floyd就好了。由于数据很大,要用long long。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#pragma GCC optimize("Og")
#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))
typedef long long ll;
const ll INF=LONG_LONG_MAX;
const int MAXN=100+10;
using namespace std;
int n,m;
ll edge[MAXN][MAXN];
int main()
{
    int T,cas=1;
    cin>>T;
    while(T--)
    {
        ll l1,l2,l3,l4,c1,c2,c3,c4;
        cin>>l1>>l2>>l3>>l4>>c1>>c2>>c3>>c4;
        ll cz[MAXN];
        cin>>n>>m;
        for(int i=1; i<=n; i++)
            cin>>cz[i];
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
            {
                ll cost=abs(cz[i]-cz[j]);
                if(cost<=l1)edge[i][j]=c1;
                else if(cost<=l2)edge[i][j]=c2;
                else if(cost<=l3)edge[i][j]=c3;
                else if(cost<=l4)edge[i][j]=c4;
                else edge[i][j]=INF;
            }
        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&&edge[i][k]+edge[k][j]<edge[i][j])
                        edge[i][j]=edge[i][k]+edge[k][j];
        printf("Case %d:\n",cas++);
        for(int i=0; i<m; i++)
        {
            int u,v;
            cin>>u>>v;
            if(edge[u][v]<INF)
            {
                printf("The minimum cost between station %d and station %d is ",u,v);
                cout<<edge[u][v]<<"."<<endl;
            }
            else printf("Station %d and station %d are not attainable.\n",u,v);
        }
    }
    return 0;
}

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