B. Cells Not Under Attack
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 
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.

题意:n*n的棋盘  m个rook(车 象棋中的ju)  给你m的车的坐标 车所在的行列不能摆放棋子

每摆放一个车 输出还能摆放其他旗子的位置的数量  具体看样例

题解:每次增加一个车 都更新不能摆放棋子的行列的个数 ro,cl

注意行列的交叉点会重复计算 记录每次的结果为ans=n*n-(ro+cl)*n+ro*cl;

 //code  by drizzle
#include<bits/stdc++.h>
#include<iostream>
#include<cstring>
#include<cstdio>
#define ll __int64
#define PI acos(-1.0)
#define mod 1000000007
using namespace std;
ll n,m;
ll h[];
ll l[];
ll cl,ro;
ll x,y;
ll ans[];
int main()
{
cl=;
ro=;
scanf("%I64d %I64d",&n,&m);
memset(l,,sizeof(l));
memset(h,,sizeof(h));
for(int i=;i<=m;i++)
{
scanf("%I64d %I64d",&x,&y);
if(h[x]==)
{
h[x]=;
ro++;
}
if(l[y]==)
{
l[y]=;
cl++;
}
ans[i]=n*n-(ro+cl)*n+ro*cl;
}
printf("%I64d",ans[]);
for(int i=;i<=m;i++)
printf(" %I64d",ans[i]);
return ;
}

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

  1. 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, ...

  2. 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 ...

  3. 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 ...

  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) C 二分处理+求区间不同字符的个数 尺取法

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

  6. 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 ...

  7. 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: ...

  8. 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 ...

  9. 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 ...

随机推荐

  1. SpringBoot学习1:创建第一个SpringBoot项目

    一.新建项目 二.打开项目的pom文件,在里面添加maven依赖 <!--springboot项目依赖的父项目--> <parent> <groupId>org.s ...

  2. docker部署Ceph分布式存储集群

    1.环境准备 3台virtualbox虚拟机,用来安装ceph集群,已用docker-machine安装上了docker,每台虚拟机虚拟创建一个5G的硬盘,用于存储osd数据,例如:/dev/sdb ...

  3. 微信小游戏 demo 飞机大战 代码分析 (一)(game.js, main.js)

    微信小游戏 demo 飞机大战 代码分析(一)(main.js) 微信小游戏 demo 飞机大战 代码分析(二)(databus.js) 微信小游戏 demo 飞机大战 代码分析(三)(spirit. ...

  4. django1.11文档 模型重点笔记

    模型最重要的属性是Manager. 它是Django 模型进行数据库查询操作的接口,并用于从数据库提取实例. 如果没有自定义Manager,则默认的名称为objects. Managers 只能通过模 ...

  5. 使用Navicat连接阿里云ECS服务器上的MySQL数据库

    一.首先要mysql授权 mysql>GRANT ALL PRIVILEGES ON *.* TO 'root'@'%' IDENTIFIED BY '你的mysql数据库密码' WITH GR ...

  6. 爬虫之Scarpy.Request

    一 .Request 1.request Scarpy中的HTTP请求对象 1.1.Requse的构造 #我们ctrl+左键可以看到Scarpy.Request的代码 class Request(ob ...

  7. Base64及其Python实现

    1. 什么是Base64 Base64是一种基于64个可打印字符来表示二进制数据的表示方法 Base64是一种编码方式,提及编码方式,必然有其对应的字符集合.在Base64编码中,相互映射的两个集合是 ...

  8. Admin站点

    使用admin站点 a.在settings.py中设置语言和时区 LANGUAGE_CODE = 'zh-hans' # 使用中国语言 TIME_ZONE = 'Asia/Shanghai' # 使用 ...

  9. Kubernetes添加带Quota限额的CephFS StorageClass

    1. 在Ceph上为Kubernetes创建一个文件系统 # ceph osd pool create cephfs_data # ceph osd pool create cephfs_metada ...

  10. 如何在C#中调试LINQ查询

    原文:How to Debug LINQ queries in C# 作者:Michael Shpilt 译文:如何在C#中调试LINQ查询 译者:Lamond Lu 在C#中我最喜欢的特性就是LIN ...