CF701 B. Cells Not Under Attack

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;
}
posted @ 2021-07-01 19:56  Wolfycz  阅读(19)  评论(0编辑  收藏  举报