题意很明确 给你一个图某坐标上的星星亮 暗 条件 求出当前区间内所有亮星星的总数
#include <iostream> #include<cstdio> #include<cstring> using namespace std; const int maxn = 1005; int tree[maxn + 2][maxn + 2]; inline int Lowbit(int x) { return x&(-x); } inline void Update(int x,int y,int v) { for(int i=x; i<=maxn; i+=Lowbit(i)) for(int j=y; j<=maxn; j+=Lowbit(j)) tree[i][j]+=v; } inline int Query(int x,int y) { int sum=0; for(int i=x; i>0; i-=Lowbit(i)) for(int j=y; j>0; j-=Lowbit(j)) sum+=tree[i][j]; return sum; } bool map[1005][1005]; int main() { int n; char c[3]; memset(tree,0,sizeof(tree)); memset(map,0,sizeof(map)); cin>>n; while(n--) { scanf("%s",c); if(c[0]=='B') { int x,y; scanf("%d%d",&x,&y); if(!map[x+1][y+1]) map[x+1][y+1]=1,Update(x+1,y+1,1); } else if(c[0]=='D') { int x,y; scanf("%d%d",&x,&y); if(map[x+1][y+1]==1) map[x+1][y+1]=0,Update(x+1,y+1,-1); } else if(c[0]=='Q') { int x1,x2,y1,y2; scanf("%d%d%d%d",&x1,&x2,&y1,&y2); if(x1>x2) swap(x1,x2); if(y1>y2) swap(y1,y2); int ans=Query(x2+1,y2+1)-Query(x1,y2+1)-Query(x2+1,y1)+Query(x1,y1); printf("%d\n",ans); } } return 0; }