codeforces 701B B. Cells Not Under Attack(水题)

题目链接：

B. Cells Not Under Attack

题意：

n*n的棋盘,现在放m个棋子,放一个棋子这一行和这一列就不会under attack了,每次放棋子回答有多少点还可能under attack;

思路：对行和列标记;

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e5+10;
const int maxn=500+10;
const double eps=1e-6;

int n,x[N],y[N];

int main()
{       
        int n,m;
        read(n);read(m);
        LL ans=(LL)n*n;
        int fx=0,fy=0;
        For(i,1,m)
        {
            int a,b;
            read(a);read(b);
            if(!x[a])
            {
                ans=ans-(n-fy);
                x[a]=1;
                fx++;
            }
            if(!y[b])
            {
                ans=ans-(n-fx);
                y[b]=1;
                fy++;
            }
            cout<<ans<<" ";
        } 
        return 0;
}