Description

Vasya has the square chessboard of size n × n and m rooks. Initially the chessboard is empty. Vasya will consequently put the rooks on the board one after another.

The cell of the field is under rook's attack, if there is at least one rook located in the same row or in the same column with this cell. If there is a rook located in the cell, this cell is also under attack.

You are given the positions of the board where Vasya will put rooks. For each rook you have to determine the number of cells which are not under attack after Vasya puts it on the board.

Input

The first line of the input contains two integers n and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ min(100 000, n2)) — the size of the board and the number of rooks.

Each of the next m lines contains integers xi and yi (1 ≤ xi, yi ≤ n) — the number of the row and the number of the column where Vasya will put the i-th rook. Vasya puts rooks on the board in the order they appear in the input. It is guaranteed that any cell will contain no more than one rook.

Output

Print m integer, the i-th of them should be equal to the number of cells that are not under attack after first i rooks are put.

Examples
input
3 3
1 1
3 1
2 2
output
4 2 0 
input
5 2
1 5
5 1
output
16 9 
input
100000 1
300 400
output
9999800001 

我们把x,y记录一下,根据情况去行还是去列
#include<bits/stdc++.h>
using namespace std;
long long n,m;
long long sum;
long long px[100005],py[100005];
long long ans[100005];
int main()
{
long long x,y;
int pos;
cin>>n>>m;
pos=n;
sum=n*n;
for(int i=1; i<=m; i++)
{
cin>>x>>y;
if(px[x]==0&&py[y]==0)
{
n--;
pos--;
// ans[i]=n*pos;
// cout<<"A"<<endl;
}
else if(px[x]&&py[y]==0)
{
pos--;
// ans[i]=n*pos;
}
else if(px[x]==0&&py[y])
{
n--;
// ans[i]=n*pos;
}
px[x]=1;
py[y]=1;
cout<<pos*n<<endl;
}
return 0;
}

  

Note

On the picture below show the state of the board after put each of the three rooks. The cells which painted with grey color is not under the attack.

Codeforces Round #364 (Div. 2) B的更多相关文章

  1. Codeforces Round #364 (Div. 2)

    这场是午夜场,发现学长们都睡了,改主意不打了,第二天起来打的virtual contest. A题 http://codeforces.com/problemset/problem/701/A 巨水无 ...

  2. Codeforces Round #364 (Div.2) D:As Fast As Possible(模拟+推公式)

    题目链接:http://codeforces.com/contest/701/problem/D 题意: 给出n个学生和能载k个学生的车,速度分别为v1,v2,需要走一段旅程长为l,每个学生只能搭一次 ...

  3. Codeforces Round #364 (Div.2) C:They Are Everywhere(双指针/尺取法)

    题目链接: http://codeforces.com/contest/701/problem/C 题意: 给出一个长度为n的字符串,要我们找出最小的子字符串包含所有的不同字符. 分析: 1.尺取法, ...

  4. 树形dp Codeforces Round #364 (Div. 1)B

    http://codeforces.com/problemset/problem/700/B 题目大意:给你一棵树,给你k个树上的点对.找到k/2个点对,使它在树上的距离最远.问,最大距离是多少? 思 ...

  5. Codeforces Round #364 (Div. 2) B. Cells Not Under Attack

    B. Cells Not Under Attack time limit per test 2 seconds memory limit per test 256 megabytes input st ...

  6. Codeforces Round #364 (Div. 2) Cells Not Under Attack

    Cells Not Under Attack 题意: 给出n*n的地图,有给你m个坐标,是棋子,一个棋子可以把一行一列都攻击到,在根据下面的图,就可以看出让你求阴影(即没有被攻击)的方块个数 题解: ...

  7. Codeforces Round #364 (Div. 2) Cards

    Cards 题意: 给你n个牌,n是偶数,要你把这些牌分给n/2个人,并且让每个人的牌加起来相等. 题解: 这题我做的时候,最先想到的是模拟,之后码了一会,发现有些麻烦,就想别的方法.之后发现只要把它 ...

  8. Codeforces Round #364 (Div. 2)->A. Cards

    A. Cards time limit per test 1 second memory limit per test 256 megabytes input standard input outpu ...

  9. Codeforces Round #364 (Div. 2) E. Connecting Universities

    E. Connecting Universities time limit per test 3 seconds memory limit per test 256 megabytes input s ...

  10. Codeforces Round #364 (Div. 2) C.They Are Everywhere

    C. They Are Everywhere time limit per test 2 seconds memory limit per test 256 megabytes input stand ...

随机推荐

  1. 2017-2018-1 20179203《Linux内核原理与分析》第二周作业

    攥写人:李鹏举 学号:20179203 ( 原创作品转载请注明出处) ( 学习课程:<Linux内核分析>MOOC课程http://mooc.study.163.com/course/US ...

  2. Servlet的生命周期以及简单工作原理的讲解

    Servlet生命周期分为三个阶段: 1,初始化阶段              调用init()方法 2,响应客户请求阶段 调用service()方法 3,终止阶段           调用destr ...

  3. Hibernate---Hql查询2---

    hibernate.cfg.xml配置: <?xml version='1.0' encoding='UTF-8'?> <!DOCTYPE hibernate-configurati ...

  4. windows修改远程桌面RDP连接数

    windows 2003在默认情况下最多只允许两个用户进行远程终端连接,当达到两个远程桌面连接的到时候,再有人尝试连接,就会提示已经达到最大终端数,无法连上了. 一.windows2003终端连接数修 ...

  5. Mybaits整合Spring自动扫描 接口,Mybaits配置文件.xml文件和Dao实体类

    1.转自:https://blog.csdn.net/u013802160/article/details/51815077 <?xml version="1.0" enco ...

  6. javascript数字千分分隔符

    function thousandBitSeparator(num) { num=num.toFixed(2); return num && num .toString() .repl ...

  7. Eclipse/MyEclipse下如何Maven管理多个Mapreduce程序?(企业级水平)

    不多说,直接上干货! 如何在Maven官网下载历史版本 Eclipse下Maven新建项目.自动打依赖jar包(包含普通项目和Web项目) Eclipse下Maven新建Web项目index.jsp报 ...

  8. React中state和props的区别

    props和state都是用于描述component状态的,并且这个状态应该是与显示相关的. State 如果component的某些状态需要被改变,并且会影响到component的render,那么 ...

  9. windows 服务器安装python及其基本库

    步骤如下: 一.安装python软件: 1.进入windows服务器,从网址下载 python-3.5.4-amd64软件 到桌面: 2.在软件点右键,再“”以管理员身份运行“”,输入管理员密码: 3 ...

  10. Appium 在 Android UI 测试中的应用

    原文地址:https://blog.coding.net/blog/Appium-Android-UI Android 测试工具与 Appium 简介 Appium 是一个 C/S 架构的,支持 An ...