hdu 1659——Bus System(Floyd)

Problem Description

Because of the huge population of China, publictransportation is very important. Bus is an important transportation method intraditional public transportation system. And it’s still playing an importantrole even now.
The bus system of City X is quite strange. Unlike other city’s system, the costof ticket is calculated based on the distance between the two stations. Here isa 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 tocalculate the minimum cost between the two stations he listed. Can you solvethis problem for him?
To simplify this problem, you can assume that all the stations are located on astraight line. We use x-coordinates to describe the stations’ positions.

 

 

Input

The input consistsof 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 stationsand questions. Each of the next n lines contains one integer, representing thex-coordinate of the ith station. Each of the next m lines contains twointegers, representing the start point and the destination.
In all of the questions, the start point will be different from thedestination.
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 samevalue.

 

 

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 formatin 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 costbetween station 1 and station 4 is 3.

The minimum costbetween station 4 and station 1 is 3.

Case 2:

Station 1 andstation 4 are not attainable.

 

 

Source

2008 “SunlineCup” National Invitational Contest

 

题意：讲的一个公交系统，该系统通过起点和终点的距离进行收费，然后给出四个不同的l和c，然后是路径，然后m次询问，对于每次询问输出最小的花费！

 

思路：因为询问有多组，所以自然想到用Floyd，在n个路径确定以后，根据表中的l让他映射到不同的价格，最后用Floyd处理即可！(注意：因为数据很大，图要用long long 保存)

 

Code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long ll;
const int N=110;
const ll INF=0x7fffffffffffffff; //此处也要注意
ll mp[N][N];
ll l[8],c[8];
int n,m;

void Floyd()
{
    int i,j,k;
    for (k=0;k<n;k++)
        for (i=0;i<n;i++)
          for (j=0;j<n;j++)
            if (mp[i][k]<INF&&mp[k][j]<INF)
             mp[i][j]=min(mp[i][k]+mp[k][j],mp[i][j]);
}

int main()
{
    int T,ca;
    int i,j,k;
    scanf("%d",&T);
    for (ca=1;ca<=T;ca++)
    {
        for (i=0;i<4;i++)
        cin>>l[i];
        for (i=0;i<4;i++)
        cin>>c[i];
        scanf("%d%d",&n,&m);
        for( i=0;i<n;i++)
            for(j=0;j<n;j++)
                mp[i][j]= (i==j?0:INF);
        int x[N];
        for (i=0;i<n;i++)
            scanf("%d",&x[i]);
        for (i=0;i<n;i++)
            for (j=i+1;j<n;j++)
            {
                for (k=0;k<4;k++)
                {
                    if (abs(x[i]-x[j])<=l[k])
                    {
                        mp[i][j]=mp[j][i]=c[k];
                        k=0;
                          break;
                    }
                }
            if (k>=4) mp[i][j]=mp[j][i]=INF;
            }
        Floyd();
        //cout<<mp[0][3]<<endl;
        printf("Case %d:\n",ca);
        while (m--)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            if (mp[u-1][v-1]==INF)  printf("Station %d and station %d are not attainable.\n",u,v);
            else printf("The minimum cost between station %d and station %d is %lld.\n",u,v,mp[u-1][v-1]);
        }

    }
}