zoukankan      html  css  js  c++  java
  • POJ 2376 Cleaning Shifts

    Cleaning Shifts
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 8457   Accepted: 2263

    Description

    Farmer John is assigning some of his N (1 <= N <= 25,000) cows to do some cleaning chores around the barn. He always wants to have one cow working on cleaning things up and has divided the day into T shifts (1 <= T <= 1,000,000), the first being shift 1 and the last being shift T. 
    Each cow is only available at some interval of times during the day for work on cleaning. Any cow that is selected for cleaning duty will work for the entirety of her interval. 
    Your job is to help Farmer John assign some cows to shifts so that (i) every shift has at least one cow assigned to it, and (ii) as few cows as possible are involved in cleaning. If it is not possible to assign a cow to each shift, print -1.

    Input

    * Line 1: Two space-separated integers: N and T 
    * Lines 2..N+1: Each line contains the start and end times of the interval during which a cow can work. A cow starts work at the start time and finishes after the end time.

    Output

    * Line 1: The minimum number of cows Farmer John needs to hire or -1 if it is not possible to assign a cow to each shift.

    Sample Input

    3 10
    1 7
    3 6
    6 10

    Sample Output

    2

    Hint

    This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed. 
    INPUT DETAILS: 
    There are 3 cows and 10 shifts. Cow #1 can work shifts 1..7, cow #2 can work shifts 3..6, and cow #3 can work shifts 6..10. 
    OUTPUT DETAILS: 
    By selecting cows #1 and #3, all shifts are covered. There is no way to cover all the shifts using fewer than 2 cows.
     
    每头牛有一定的工作时间段,问最少用多少头牛才能保证在所有时间内都有牛在工作
    用贪心算法,每次选取工作结束时间最晚的牛
     
     1 #include<iostream>
     2 #include<cstdio>
     3 #include<algorithm>
     4 
     5 using namespace std;
     6 
     7 typedef struct
     8 {
     9     int Start;
    10     int End;
    11 } COW;
    12 
    13 int n,t;
    14 COW cow[25010];
    15 
    16 bool cmp(COW a,COW b)
    17 {
    18     return a.Start<b.Start||(a.Start==b.Start&&a.End<b.End);
    19 }
    20 
    21 int main()
    22 {
    23     while(scanf("%d %d",&n,&t)==2)
    24     {
    25         for(int i=0;i<n;i++)
    26             scanf("%d %d",&cow[i].Start,&cow[i].End);
    27 
    28         sort(cow,cow+n,cmp);
    29 
    30         bool flag=true;
    31         int End=0,pos=0,ans=0;
    32         while(End<t)
    33         {
    34             int maxx=-1;
    35             for(;cow[pos].Start<=End+1;pos++)
    36             {
    37                 if(cow[pos].End>maxx)
    38                     maxx=cow[pos].End;
    39             }
    40             if(maxx==-1)
    41             {
    42                 flag=false;
    43                 break;
    44             }
    45             End=maxx;
    46             ans++;
    47         }
    48         if(flag)
    49             cout<<ans<<endl;
    50         else
    51             cout<<"-1"<<endl;
    52     }
    53 
    54     return 0;
    55 }
    [C++]
  • 相关阅读:
    Delphi以及三方控件的源代码规模
    InitInheritedComponent的执行过程
    poj 3897 Maze Stretching 二分+A*搜索
    一些窗口API函数,比如SetForegroundWindow,SwitchToThisWindow
    终于懂了:WM_PAINT 与 WM_ERASEBKGND(三种情况:用户操作,UpdateWindow,InvalidateRect产生的效果并不相同),并且用Delphi代码验证 good
    窗口绘制有关的消息整理 WM_PAINT, WM_NCPAINT, WM_ERASEBKGND
    WM_PAINT与WM_ERASEBKGND(用户操作和API这两种情况产生消息的顺序有所不同)
    关于WM_ERASEBKGND和WM_PAINT的深刻理解
    offsetHeight在OnLoad中为0的现象
    TWinControl.WMNCPaint对非客户的绘制
  • 原文地址:https://www.cnblogs.com/lzj-0218/p/3246496.html
Copyright © 2011-2022 走看看