标签：des style blog class code c
Bus System
Time Limit: 2000/1000 MS
(Java/Others) Memory Limit: 32768/32768 K
(Java/Others)
Total Submission(s): 6082 Accepted
Submission(s): 1560
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
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简单最短路径，注意是64位：
1 //125MS 288K 1415B G++
2 #include<stdio.h>
3 #include<string.h>
4 #define N 105
5 #define inf 1000000000000LL
6 __int64 g[N][N];
7 int n,m;
8 __int64 abs(__int64 a)
9 {
10 return a<0?-a:a;
11 }
12 __int64 Min(__int64 a,__int64 b)
13 {
14 return a<b?a:b;
15 }
16 void floyd() //弗洛伊德算法
17 {
18 for(int k=0;k<n;k++)
19 for(int i=0;i<n;i++)
20 for(int j=0;j<n;j++)
21 g[i][j]=Min(g[i][j],g[i][k]+g[k][j]);
22 }
23 int main(void)
24 {
25 int t,k=1;
26 __int64 l1,l2,l3,l4,c1,c2,c3,c4,x[N];
27 int a,b;
28 scanf("%d",&t);
29 while(t--)
30 {
31 scanf("%I64d%I64d%I64d%I64d%I64d%I64d%I64d%I64d",&l1,&l2,&l3,&l4,&c1,&c2,&c3,&c4);
32 scanf("%d%d",&n,&m);
33 for(int i=0;i<n;i++) scanf("%I64d",&x[i]);
34 for(int i=0;i<n;i++)
35 for(int j=0;j<n;j++){
36 int L=abs(x[i]-x[j]);
37 if(L==0) g[i][j]=0;
38 else if(L<=l1) g[i][j]=c1;
39 else if(L<=l2) g[i][j]=c2;
40 else if(L<=l3) g[i][j]=c3;
41 else if(L<=l4) g[i][j]=c4;
42 else g[i][j]=inf;
43 }
44 floyd();
45 printf("Case %d:\n",k++);
46 while(m--){
47 scanf("%d%d",&a,&b);
48 if(g[a-1][b-1]==inf) printf("Station %d and station %d are not attainable.\n",a,b);
49 else printf("The minimum cost between station %d and station %d is %I64d.\n",a,b,g[a-1][b-1]);
50 }
51 }
52 return 0;
53 }
hdu 1690 Bus System (最短路径),布布扣,bubuko.com
hdu 1690 Bus System (最短路径)
标签：des style blog class code c
原文地址：http://www.cnblogs.com/GO-NO-1/p/3733475.html
