zoukankan      html  css  js  c++  java
  • Hdu

    先上题目:

    Holedox Eating

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 2793    Accepted Submission(s): 919


    Problem Description
    Holedox is a small animal which can be considered as one point. It lives in a straight pipe whose length is L. Holedox can only move along the pipe. Cakes may appear anywhere in the pipe, from time to time. When Holedox wants to eat cakes, it always goes to the nearest one and eats it. If there are many pieces of cake in different directions Holedox can choose, Holedox will choose one in the direction which is the direction of its last movement. If there are no cakes present, Holedox just stays where it is.
     
    Input
    The input consists of several test cases. The first line of the input contains a single integer T (1 <= T <= 10), the number of test cases, followed by the input data for each test case.The first line of each case contains two integers L,n(1<=L,n<=100000), representing the length of the pipe, and the number of events.
    The next n lines, each line describes an event. 0 x(0<=x<=L, x is a integer) represents a piece of cake appears in the x position; 1 represent Holedox wants to eat a cake.
    In each case, Holedox always starts off at the position 0.
     
    Output
    Output the total distance Holedox will move. Holedox don’t need to return to the position 0.
     
    Sample Input
    3 10 8 0 1 0 5 1 0 2 0 0 1 1 1 10 7 0 1 0 5 1 0 2 0 0 1 1 10 8 0 1 0 1 0 5 1 0 2 0 0 1 1
     
    Sample Output
    Case 1: 9 Case 2: 4 Case 3: 2
     
      题意是在一条直线上一开始一只小动物在0点,然后有两种操作,一种是在某一点上放一块蛋糕,另一种是去吃一块蛋糕。其中吃蛋糕有规定,先吃离小动物最近的蛋糕,如果有两块蛋糕离小动物的距离一样近,先吃当时的朝向的蛋糕,问所有操作以后小动物所走的路多长。
      这题又看就是模拟题,没有什么太难搞的地方,当然做法就是记录当前小动物的左边和右边最近的那一块蛋糕的位置,还有小动物的朝向,然后就直接模拟就可以了。当然,这里我为了方便用了优先队列来保存当前位置的左右两边的蛋糕位置。
     
     
    上代码:
     1 #include <stdio.h>
     2 #include <string.h>
     3 #include <queue>
     4 #include <math.h>
     5 #include <algorithm>
     6 #define MAX (100000+10)
     7 using namespace std;
     8 
     9 long long sum;
    10 
    11 struct cmp1
    12 {
    13     bool operator() (int x,int y)
    14     {
    15         return x<y;
    16     }
    17 };
    18 
    19 struct cmp2
    20 {
    21     bool operator() (int x,int y)
    22     {
    23         return x>y;
    24     }
    25 };
    26 
    27 priority_queue<int,vector<int>,cmp1> L;
    28 priority_queue<int,vector<int>,cmp2> R;
    29 
    30 int main()
    31 {
    32     int t,j,n,c,w,m,r;
    33     bool f;
    34     //freopen("data.txt","r",stdin);
    35     scanf("%d
    ",&t);
    36     for(j=1; j<=t; j++)
    37     {
    38         while(!L.empty()) L.pop();
    39         while(!R.empty()) R.pop();
    40         sum=0;
    41         m=0;
    42         f=1;
    43         scanf("%d %d",&r,&n);
    44         while(n--)
    45         {
    46             scanf("%d",&c);
    47             if(c)
    48             {
    49                 if(L.empty() && R.empty()) continue;
    50                 if(L.empty())
    51                 { sum+=R.top()-m; m=R.top(); R.pop(); f=1; }
    52                 else if(R.empty())
    53                 { sum+=m-L.top(); m=L.top(); L.pop(); f=0; }
    54                 else if(m-L.top()<R.top()-m)
    55                 { sum+=m-L.top(); m=L.top(); L.pop(); f=0; }
    56                 else if(R.top()-m<m-L.top())
    57                 { sum+=R.top()-m; m=R.top(); R.pop(); f=1; }
    58                 else if(f)
    59                 { sum+=R.top()-m; m=R.top(); R.pop(); f=1; }
    60                 else
    61                 { sum+=m-L.top(); m=L.top(); L.pop(); f=0; }
    62             }
    63             else
    64             {
    65                 scanf("%d",&w);
    66                 if(w<m) L.push(w);
    67                 else R.push(w);
    68             }
    69         }
    70         printf("Case %d: %lld
    ",j,sum);
    71     }
    72     return 0;
    73 }
    4302
  • 相关阅读:
    Deployment of VC2008 apps without installing anything
    用MT.exe将exe中的manifest文件提取出来和将manifest文件放入exe中
    Golang快速入门
    8个优质的编程学习网站
    免费学编程!10个全球顶尖的编程在线自学网站
    7个在线学习C++编程的最佳途径
    为什么多数游戏服务端是用 C++ 来写
    学习游戏服务器开发必看,C++游戏服务器开发常用工具介绍
    Ambari——大数据平台的搭建利器(一)
    Python爬虫项目整理
  • 原文地址:https://www.cnblogs.com/sineatos/p/3230934.html
Copyright © 2011-2022 走看看