• poj 2376 Cleaning Shifts


    http://poj.org/problem?id=2376

    Cleaning Shifts
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 12604   Accepted: 3263

    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.
     
     
    分析:
    题意就是给你N段小区间 [n , m]和一个T,T代表大区间[1 ,T] , 要求你找出最少使用小区间完全覆盖大区间。
    一开始一直WA,因为理解错了“完全覆盖”的概念。即是下面的数据:
    1     3 10      //3代表小区间个数,10代表大区间长度。
    2     1 5
    3     6 8
    4     9 10

    上面的数据我的程序输出的是:-1

    因为有两个区间没有“完全覆盖”,[5 ,6]和[8 ,9]。

    但是正确答案是 3 。

    WA代码:

     1 #include<cstdio>
     2 #include<algorithm>
     3 using namespace std;
     4 struct P
     5 {
     6     int x,y;
     7     bool operator < (const P & p) const
     8     {
     9         return x < p.x || (x == p.x && y > p.y);
    10     }
    11 }a[25010];
    12 
    13 int main()
    14 {
    15     int n,t;
    16     while(~scanf("%d %d",&n,&t))
    17     {
    18         for(int i = 0;i < n;i++) 
    19             scanf("%d %d",&a[i].x,&a[i].y);
    20         sort(a ,a + n);
    21         int res = 1,s;
    22         if(a[0].x > 1)
    23             printf("-1
    ");
    24         else
    25         {
    26             s = a[0].y;
    27             for(int i = 1;i < n && s < t;)
    28             {
    29                 int tmp = 0;
    30                 while(i < n && a[i].x <= s)
    31                 {
    32                     tmp = max(tmp , a[i].y);
    33                     i++;
    34                 }
    35                 if(tmp > s)
    36                 {
    37                     s = tmp;
    38                     res++;
    39                 }
    40                 else
    41                     break;
    42             }
    43         }
    44         if(s >= t)
    45             printf("%d
    ",res);
    46         else
    47             printf("-1
    ");
    48     }
    49     return 0;
    50 }
    View Code

    AC代码:

     1 #include<cstdio>
     2 #include<algorithm>
     3 using namespace std;
     4 
     5 struct P{
     6     int x,y;
     7     bool operator < (const P & p) const {
     8         return x < p.x || (x == p.x && y > p.y);
     9     }
    10 }a[25010];
    11 
    12 int main() {
    13     int n,t;
    14     while(~scanf("%d %d",&n,&t)) {
    15         for(int i = 0;i < n;i++) 
    16             scanf("%d %d",&a[i].x,&a[i].y);
    17             
    18         sort(a ,a + n);
    19         
    20         int res = 1,s;
    21         if(a[0].x > 1) {
    22             printf("-1
    ");
    23             continue;
    24         } else {
    25             s = a[0].y;
    26             for(int i = 1;i < n && s < t;) {
    27                 int tmp = 0;
    28                 while(i < n && a[i].x <= s + 1) {
    29                     tmp = max(tmp , a[i].y);
    30                     i++;
    31                 }
    32                 if(tmp > s) {
    33                     s = tmp;
    34                     res++;
    35                 } else break;
    36             }
    37         }
    38             
    39         if(s >= t) {
    40             printf("%d
    ",res);
    41         } else {
    42             printf("-1
    ");
    43         }
    44     }
    45     return 0;
    46 }
  • 相关阅读:
    zookeeperclient代码解读
    封装scrollView 循环滚动,tableViewCell(连载) mvc
    PHP经典项目案例-(一)博客管理系统5
    Android插件化(三)载入插件apk中的Resource资源
    比树莓派配置好接地气的香蕉派上手初体验
    HDU Group
    JVM 类的卸载
    JVM 自定义类加载器
    JVM 初始化阶段例子
    JVM 初始化阶段例子 final常量
  • 原文地址:https://www.cnblogs.com/jeff-wgc/p/4478165.html
Copyright © 2020-2023  润新知