题目链接:
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(‘\n‘); } 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; }
时间: 2024-10-10 15:59:46