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 #include #include #include #include #include #include #define Fi first #define Se second #define ll_inf 1e18 #define MK make_pair #define sqr(x) ((x)*(x)) #define pii pair #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++; } templateinline 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; } templateinline 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; }