hdu Bus System（floyd）

Bus System

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4239    Accepted Submission(s): 1066

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

21 2 3 4 1 3 5 74 212341 44 11 2 3 4 1 3 5 74 1123101 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

 

Recommend

wangye

因为这道题的询问m>100，所以用floyd比较好

开始把不能联通的路权值设为-1，结果不停的wa

以为-1和Max是一样的效果，都只是一个标记，后来问了喵呜大神，结果被他一眼看出错误，orz。。。

详见代码

#include <stdio.h>#define Max 0xfffffffffffint x[105];__int64 cost[105][105];int main(){    int t,i,j,k,q;    int l1,l2,l3,l4,c1,c2,c3,c4;    int n,m;    int s,d;    scanf("%d",&t);    for(q = 1;q <= t;q++){        scanf("%d%d%d%d%d%d%d%d",&l1,&l2,&l3,&l4,&c1,&c2,&c3,&c4);        scanf("%d%d",&n,&m);        for(i = 1;i <= n;i++)            scanf("%d",&x[i]);        for(i = 1;i <= n;i++)            for(j = 1;j <= n;j++){                int der = x[i] - x[j];                if(der < 0) der = -der;                if(der == 0)        cost[i][j] = 0;                else if(der <= l1)  cost[i][j] = c1;                else if(der <= l2)  cost[i][j] = c2;                else if(der <= l3)  cost[i][j] = c3;                else if(der <= l4)  cost[i][j] = c4;                else                cost[i][j] = Max;            }        for(k = 1;k <= n;k++)            for(i = 1;i <= n;i++)                for(j = 1;j <= n;j++){//如果Max设为-1，则某初始无法连通的两点，在经过floyd后仍无法连通，因为-1比所有cost都要小                    if(cost[i][k] != Max && cost[k][j] != Max && cost[i][k] + cost[k][j] < cost[i][j])                        cost[i][j] = cost[i][k] + cost[k][j];                }        printf("Case %d:\n",q);        while(m--){            scanf("%d%d",&s,&d);            if(cost[s][d] == Max)                printf("Station %d and station %d are not attainable.\n",s,d);            else                printf("The minimum cost between station %d and station %d is %I64d.\n",s,d,cost[s][d]);        }    }}