题目传送门:https://codeforces.com/problemset/problem/701/B
题目大意:
给定一个\(n\times n\)的棋盘,共有\(m\)次操作,每次操作会在棋盘的\((x,y)\)处放置一个城堡(国际象棋),问每次操作后不会被攻击的格子总数
每次放置城堡只会影响到一整行和一整列,故我们用两个数组记录行和列的放置情况
每次加入新的城堡的时候进行判断即可
/*program from Wolfycz*/
#include<map>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define Fi first
#define Se second
#define ll_inf 1e18
#define MK make_pair
#define sqr(x) ((x)*(x))
#define pii pair<int,int>
#define int_inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
static char buf[1000000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
template<typename T>inline T frd(T x){
int f=1; char ch=gc();
for (;ch<'0'||ch>'9';ch=gc()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=gc()) x=(x<<1)+(x<<3)+ch-'0';
return x*f;
}
template<typename T>inline T read(T x){
int f=1; char ch=getchar();
for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0';
return x*f;
}
inline void print(int x){
if (x<0) putchar('-'),x=-x;
if (x>9) print(x/10);
putchar(x%10+'0');
}
const int N=1e5;
bool VisR[N+10],VisC[N+10];
//R(Row)表示行的占用情况,C(Column)表示列的占用情况
int main(){
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
int n=read(0),m=read(0);
int totR=n,totC=n; ll All=1ll*n*n;
//totR,totC表示剩余未放置城堡的行数和列数
for (int i=1;i<=m;i++){
int x=read(0),y=read(0);
if (VisR[x]&&VisC[y]){//行和列都已放置了城堡
printf("%lld\n",All);
continue;
}
if (!VisC[y]) All-=totR;
if (!VisR[x]) All-=totC;
if (!(VisR[x]^VisC[y])) All++; //(x,y)算重,得减去一次
totR-=!VisR[x],totC-=!VisC[y];
VisR[x]=VisC[y]=1;
printf("%lld%c",All,i==m?'\n':' ');
}
return 0;
}