zoukankan      html  css  js  c++  java
  • HDU1050(贪心水题)

    Description

    The famous ACM (Advanced Computer Maker) Company has rented a floor of a building whose shape is in the following figure.



    The floor has 200 rooms each on the north side and south side along the corridor. Recently the Company made a plan to reform its system. The reform includes moving a lot of tables between rooms. Because the corridor is narrow and all the tables are big, only one table can pass through the corridor. Some plan is needed to make the moving efficient. The manager figured out the following plan: Moving a table from a room to another room can be done within 10 minutes. When moving a table from room i to room j, the part of the corridor between the front of room i and the front of room j is used. So, during each 10 minutes, several moving between two rooms not sharing the same part of the corridor will be done simultaneously. To make it clear the manager illustrated the possible cases and impossible cases of simultaneous moving.



    For each room, at most one table will be either moved in or moved out. Now, the manager seeks out a method to minimize the time to move all the tables. Your job is to write a program to solve the manager’s problem.
     

    Input

    The input consists of T test cases. The number of test cases ) (T is given in the first line of the input. Each test case begins with a line containing an integer N , 1<=N<=200 , that represents the number of tables to move. Each of the following N lines contains two positive integers s and t, representing that a table is to move from room number s to room number t (each room number appears at most once in the N lines). From the N+3-rd line, the remaining test cases are listed in the same manner as above.
     

    Output

    The output should contain the minimum time in minutes to complete the moving, one per line.
     

    Sample Input

    3 4 10 20 30 40 50 60 70 80 2 1 3 2 200 3 10 100 20 80 30 50
     

    Sample Output

    10 20 30
     
    简单的贪心,但做这道题时却一直错,我是结束后才发现问题,原来是从一开始,有些无语,一道水题,简单说一下:
    从第一个开始寻找和他可以同时完成的,标记最后计算有多少个不能同时完成的组数。
     1 #include<stdio.h>
     2 #include<algorithm>
     3 #include<string.h>
     4 using namespace std;
     5 struct time
     6 {
     7     int start;
     8     int over;
     9 } a[210];
    10 int cmp(time c,time b)
    11 {
    12     return c.start<b.start;
    13 }
    14 int vis[210];
    15 int main()
    16 {
    17     int t;
    18     scanf("%d",&t);
    19     while(t--)
    20     {
    21         memset(vis,0,sizeof(vis));
    22         int n;
    23         scanf("%d",&n);
    24         for(int i=0; i<n; i++)
    25             scanf("%d%d",&a[i].start,&a[i].over);
    26         for(int i=0;i<n;i++)
    27         {
    28             if(a[i].start>a[i].over)
    29             {
    30                 int y=a[i].start;
    31                 a[i].start=a[i].over;
    32                 a[i].over=y;
    33             }
    34             if(a[i].start%2==1)
    35             a[i].start=a[i].start/2+1;
    36             else a[i].start/=2;
    37             if(a[i].over%2==1)
    38                 a[i].over=a[i].over/2+1;
    39             else
    40             a[i].over/=2;
    41         }
    42         sort(a,a+n,cmp);
    43         int step=0;
    44         int temp;
    45         for(int i=0; i<n; i++)
    46         {
    47 
    48             temp=a[i].over;
    49             if(vis[i]==0)
    50             {
    51                 step++;
    52                 for(int j=0; j<n; j++)
    53                 {
    54                     if(a[j].start>temp&&vis[j]==0)
    55                     {
    56                         temp=a[j].over;
    57                         vis[j]=1;
    58                     }
    59                 }
    60             }
    61         }
    62         printf("%d
    ",step*10);
    63     }
    64 }
  • 相关阅读:
    Centos7下安装Oracle11g r2图形化界面数据库
    power designer 16.5 使用总结[转]
    mybatis-plus忽略映射字段
    mybatis-plus快速入门使用
    git本地项目代码上传至码云远程仓库总结【转】
    北京Java笔试题整理
    linux下启动和关闭tomcat服务的方式
    SpringMvc支持跨域访问,Spring跨域访问,SpringMvc @CrossOrigin 跨域[转]
    mybatis中的#和$的区别
    Spring官网下载dist.zip的几种方法
  • 原文地址:https://www.cnblogs.com/VectorLin/p/5216650.html
Copyright © 2011-2022 走看看