Stall Reservations
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 11002   Accepted: 3886   Special Judge

Description

Oh those picky N (1 <= N <= 50,000) cows! They are so picky that each one will only be milked over some precise time interval A..B (1 <= A <= B <= 1,000,000), which includes both times A and B. Obviously, FJ must create a reservation system to determine which stall each cow can be assigned for her milking time. Of course, no cow will share such a private moment with other cows.

Help FJ by determining:

  • The minimum number of stalls required in the barn so that each cow can have her private milking period
  • An assignment of cows to these stalls over time

Many answers are correct for each test dataset; a program will grade your answer.

Input

Line 1: A single integer, N

Lines 2..N+1: Line i+1 describes cow i's milking interval with two space-separated integers.

Output

Line 1: The minimum number of stalls the barn must have.

Lines 2..N+1: Line i+1 describes the stall to which cow i will be assigned for her milking period.

Sample Input

5
1 10
2 4
3 6
5 8
4 7

Sample Output

4
1
2
3
2
4

Hint

Explanation of the sample:

Here's a graphical schedule for this output:

Time     1  2  3  4  5  6  7  8  9 10

Stall 1 c1>>>>>>>>>>>>>>>>>>>>>>>>>>>

Stall 2 .. c2>>>>>> c4>>>>>>>>> .. ..

Stall 3 .. .. c3>>>>>>>>> .. .. .. ..

Stall 4 .. .. .. c5>>>>>>>>> .. .. ..

Other outputs using the same number of stalls are possible.

Source

 
题意:人去挤奶牛,一个人挤一头,输出最少几人及牛奶并且哪只牛哪个人
题解:贪心加优先队列,将开始时间在前面的排在前面,开始时间一样的就把结束时间后的排在后面

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<map>
#include<set>
using namespace std;
#define INF 0x3f3f3f3f
const int maxn=;
int n,use[maxn]; struct node
{
int st;
int en;
int pos;
bool operator < (const node &a)const
{
if(en==a.en)
return st>a.st;
return en>a.en;
}
}a[maxn];
priority_queue<node>q;
bool cmp(node a,node b)
{
if(a.st==b.st)
return a.en<b.en; //排序,按照开始时间排序,开始时间相同的结束时间迟的排后面
return a.st<b.st;
} int main()
{
while(scanf("%d",&n)!=EOF)
{
for(int i=;i<n;i++)
{
scanf("%d %d",&a[i].st,&a[i].en);
a[i].pos=i;
}
sort(a,a+n,cmp);
q.push(a[]);
int ans=;
use[a[].pos]=;
for(int i=;i<n;i++)
{
if(!q.empty() && q.top().en < a[i].st)
{
use[a[i].pos]=use[q.top().pos];
q.pop();
}
else
{
ans++;
use[a[i].pos]=ans;
}
q.push(a[i]);
}
cout<<ans<<endl;
for(int i=;i<n;i++)
cout<<use[i]<<endl;
while(!q.empty())
q.pop();
}
return ;
}

最新文章

  1. Mac读取Andriod屏幕截图
  2. DataTables 自定义
  3. WebSocket 学习笔记--IE,IOS,Android等设备的兼容性问题与代码实现
  4. 为 Node.js 开发者准备的 8 本免费在线电子书(转)
  5. Guava 的学习
  6. linux命令(5)文件操作:ls命令、显示文件总个数
  7. dotnet core多平台开发体验
  8. JetBrains发布了一款免费的.NET反编译器dotPeek
  9. [总结]Map: C++ V.S. Java
  10. Ubuntu Codeblocks Installation
  11. Core Animation 文档翻译 (第八篇)—提高动画的性能
  12. 「HNOI 2019」白兔之舞
  13. 关于Shader的学习记录
  14. 分布式文件系统HDFS,大数据存储实战(一)
  15. 启动软件丢失 MSVCR100.dll 系列,缺少库的问题
  16. Android开发随笔记_1
  17. html5-颜色的表示
  18. day 54 JS 之 jquery
  19. loj#121.「离线可过」动态图连通性
  20. 170707、springboot编程之监控和管理生产环境

热门文章

  1. NetCore组件
  2. js和jq中常见的各种位置距离之offsetLeft和position().left的区别(四)
  3. linux 安装jdk (二进制文件安装)
  4. JavaScript中登录名的正则表达式及解析(0基础)
  5. Kendo UI 单页面应用(二) Router 类
  6. gitlab api批量操作 批量添加用户
  7. python中的构造函数和构造函数和析构函数的作用
  8. Delphi7使用二维码
  9. Arria II GX FPGA开法套件——初步使用
  10. ABAP Development Tools的语法高亮实现原理