Bus System

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8055    Accepted Submission(s): 2121

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.

注意这题INF要很大。。我就是因为INF开小了一点点就WA了。

#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <iostream>
#include <stdlib.h>
using namespace std;
typedef long long ll;
const ll INF =111111111111;
const int N = 105;
ll graph[N][N];
ll L[4],C[4];
ll x[N];
ll check(ll dis){
    if(dis<=L[0]) return C[0];
    else if(dis>L[0]&&dis<=L[1]) return C[1];
    else if(dis>L[1]&&dis<=L[2]) return C[2];
    else if(dis>L[2]&&dis<=L[3]) return C[3];
    return INF;
}
int main()
{
    int tcase;
    scanf("%d",&tcase);
    int t = 1;
    while(tcase--){
        for(int i=0;i<4;i++){
            scanf("%lld",&L[i]);
        }
        sort(L,L+4);
        for(int i=0;i<4;i++){
            scanf("%lld",&C[i]);
        }
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++){
            scanf("%lld",&x[i]);
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(i==j) graph[i][j] = 0;
                else{
                    ll dis = abs(x[i]-x[j]);
                    graph[i][j]=check(dis);
                }
            }
        }
        for(int k=0;k<n;k++){
            for(int i=0;i<n;i++){
                if(graph[k][i]<INF){
                    for(int j=0;j<n;j++){
                        graph[i][j] = min(graph[i][j],graph[i][k]+graph[k][j]);
                    }
                }
            }
        }
        printf("Case %d:\n",t++);
        while(m--){
            int a,b;
            scanf("%d%d",&a,&b);
            int c = a-1,d=b-1;
            if(graph[c][d]>=INF) printf("Station %d and station %d are not attainable.\n",a,b);
            else printf("The minimum cost between station %d and station %d is %lld.\n",a,b,graph[c][d]);
        }
    }
}