codeforces #364b Cells Not Under Attack

最新推荐文章于 2022-07-13 02:01:45 发布

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最新推荐文章于 2022-07-13 02:01:45 发布

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原文链接：http://www.cnblogs.com/superxuezhazha/p/5698088.html

本文解析了CellsNotUnderAttack问题，介绍了如何计算放置棋盘上的车后不受攻击的单元格数量。通过跟踪已放置的车对行和列的影响，避免重复计算，实现了高效的解决方案。

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

output

4 2 0

input

5 2
1 5
5 1

output

16 9

input

100000 1
300 400

output

9999800001

#include<iostream>
#include<stdio.h>
using namespace std;
int main(){
    int n,m;
    scanf("%d%d",&n,&m);
    long long sum=n;
    sum*=sum;
    int hang=0;
    int lie=0;
    int phang[100005];
    int plie[100005];
    for(int i=0;i<=n+5;i++){
        phang[i]=plie[i]=0;
    }
    long long goji=0;
    for(int i=1;i<m;i++)
    {
        int tmp1,tmp2;
        scanf("%d%d",&tmp1,&tmp2);
        if(phang[tmp1]==0&&plie[tmp2]==0){
            goji+=(n+n-1);
            goji-=lie;
            goji-=hang;
        }else if(phang[tmp1]!=0&&plie[tmp2]==0){
            goji+=(n-1);
            goji-=(hang-1);
        }else if(phang[tmp1]==0&&plie[tmp2]!=0){
            goji+=(n-1);
            goji-=(lie-1);
        }
        if(phang[tmp1]==0){
            hang++;
        }
        if(plie[tmp2]==0){
            lie++;
        }
        phang[tmp1]++;
        plie[tmp2]++;
        printf("%I64d ",sum-goji);
           
    }

    int tmp1,tmp2;
        scanf("%d%d",&tmp1,&tmp2);
        if(phang[tmp1]==0&&plie[tmp2]==0){
            goji+=((n*2)-1);
            goji-=lie;
            goji-=hang;
        }else if(phang[tmp1]!=0&&plie[tmp2]==0){
            goji+=(n-1);
            goji-=(hang-1);
        }else if(phang[tmp1]==0&&plie[tmp2]!=0){
            goji+=(n-1);
            goji-=(lie-1);
        }
        if(phang[tmp1]==0){
            hang++;
        }
        if(plie[tmp2]==0){
            lie++;
        }
        phang[tmp1]++;
        plie[tmp2]++;
        printf("%I64d\n",sum-goji);

        return 0;
}