zoukankan      html  css  js  c++  java
  • HDU 1690 Bus System

    Bus System

    Time Limit: 1000ms
    Memory Limit: 32768KB
    This problem will be judged on HDU. Original ID: 1690
    64-bit integer IO format: %I64d      Java class name: Main
     
     
    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

     
    解题:Floyd算法
     
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <cmath>
     5 #include <algorithm>
     6 #include <climits>
     7 #include <vector>
     8 #include <queue>
     9 #include <cstdlib>
    10 #include <string>
    11 #include <set>
    12 #include <stack>
    13 #define LL long long
    14 #define pii pair<int,int>
    15 #define INF 0x3f3f3f3f3f3fLL
    16 using namespace std;
    17 LL L[4],C[4],n,m;
    18 LL x[110],mp[110][110];
    19 LL calc(int i,int j){
    20     LL tmp = x[i] - x[j];
    21     if(tmp < 0) tmp = -tmp;
    22     if(tmp > 0 && tmp <= L[0]) return C[0];
    23     if(tmp > L[0] && tmp <= L[1]) return C[1];
    24     if(tmp > L[1] && tmp <= L[2]) return C[2];
    25     if(tmp > L[2] && tmp <= L[3]) return C[3];
    26     return INF;
    27 }
    28 int main() {
    29     int t,i,j,k,u,v,cs = 1;
    30     scanf("%d",&t);
    31     while(t--){
    32         for(i = 0; i < 4; i++) scanf("%I64d",L+i);
    33         for(i = 0; i < 4; i++) scanf("%I64d",C+i);
    34         scanf("%I64d %I64d",&n,&m);
    35         for(i = 1; i <= n; i++) scanf("%I64d",x+i);
    36         for(i = 1; i <= n; i++){
    37             mp[i][i] = INF;
    38             for(j = i+1; j <= n; j++){
    39                 mp[i][j] = mp[j][i] = calc(i,j);
    40             }
    41         }
    42         for(k = 1; k <= n; k++){
    43             for(i = 1; i <= n; i++){
    44                 for(j = 1; j <= n; j++){
    45                     if(mp[i][j] > mp[i][k]+mp[k][j])
    46                         mp[i][j] = mp[i][k] + mp[k][j];
    47                 }
    48             }
    49         }
    50         printf("Case %d:
    ",cs++);
    51         while(m--){
    52             scanf("%d %d",&u,&v);
    53             if(mp[u][v] >= INF){
    54                 printf("Station %d and station %d are not attainable.
    ",u,v);
    55             }else{
    56                 printf("The minimum cost between station %d and station %d is %I64d.
    ",u,v,mp[u][v]);
    57             }
    58         }
    59     }
    60     return 0;
    61 }
    View Code
  • 相关阅读:
    js 实现树效果
    (转)JavaScript 冒泡实例与阻止冒泡方法
    (转)js全页面刷新方法
    js 右键菜单
    (转)js,jQuery屏蔽鼠标右与jquery 鼠标右键事件、左键单击事件判定
    Flex取xml文件中的值
    oracle相关时间计算,得到季度第一天、最后一天
    (转)ASP.NET MVC VS2010中更改默认调试浏览器
    JS 与 后台如何获取 Cookies
    js 弹出新页面
  • 原文地址:https://www.cnblogs.com/crackpotisback/p/3945397.html
Copyright © 2011-2022 走看看