hdu 1690 Bus System【floyd】

Bus System

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7908    Accepted Submission(s): 2068

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.
 

 

Source

2008 “Sunline Cup” National Invitational Contest

 

题目大意：t组样例，给出八个值，L1，L2，L3，L4，C1,C2，C3，C4,分别对应图表中的含义，接下来两个值，n，m，一共n个点m个询问，接下来n行， 表示各个节点x坐标。m个询问两个点之间的最小花费。

这个题两个点之间一定是最短的，我们要求的是最小花费，所以我们对花费进行floyd算法的实现，注意几个点：

1、数据比较大，该用int64的就别用int，会wa。

2、输出的地方别拉了东西。

3、初始化要足够大，我们这里花费的取值范围比较大， 所以初始化也一定要足够大才能够AC。

AC代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<stdlib.h>
#include<climits>
using namespace std;
#define ll __int64
#define inf 1e18
ll map[105][105];
ll x[105];
int  n,m;
ll L1, L2, L3, L4, C1, C2, C3, C4;
void init()
{
    for(int  i=1;i<=n;i++)
    {
        for(int  j=1;j<=n;j++)
        {
            map[i][j]=inf;
        }
    }
}
void floyd()
{
    for(int i=1;i<=n;i++)
    {
        for(int  j=1;j<=n;j++)
        {
            for(int  k=1;k<=n;k++)
            {
                map[j][k]=min(map[j][k],map[j][i]+map[i][k]);
            }
        }
    }
}
ll  getscore(ll dis)
{
    if(dis>0&&dis<=L1)return C1;
    if(dis>L1&&dis<=L2)return C2;
    if(dis>L2&&dis<=L3)return C3;
    if(dis>L3&&dis<=L4)return C4;
    return inf;
}
int  main()
{
    int  kase=0;
    int  t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d%I64d%I64d%I64d%I64d%I64d%I64d%I64d",&L1,&L2,&L3,&L4,&C1,&C2,&C3,&C4);
        scanf("%d%d",&n,&m);
        init();
        for(int  i=1;i<=n;i++)
        {
            scanf("%I64d",&x[i]);
        }
        for(int  i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                map[i][j]=map[j][i]=getscore((abs)(x[i]-x[j]));
            }
        }
        floyd();
        //printf("%I64d\n",map[1][4]);
        printf("Case %d:\n",++kase);
        while(m--)
        {
            int  u,v;
            scanf("%d%d",&u,&v);
            if(map[u][v]!=inf)
            printf("The minimum cost between station %d and station %d is %I64d.\n",u,v,map[u][v]);
            if(map[u][v]==inf)
            printf("Station %d and station %d are not attainable.\n",u,v);
        }
    }
}