Stars

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/65536 K (Java/Others) Total Submission(s): 1718    Accepted Submission(s): 725 Problem Description Yifenfei is a romantic guy and he likes to count the stars in the sky. To make the problem easier,we considerate the sky is a two-dimension plane.Sometimes the star will be bright and sometimes the star will be dim.At first,there is no bright star in the sky,then some information will be given as "B x y" where 'B' represent bright and x represent the X coordinate and y represent the Y coordinate means the star at (x,y) is bright,And the 'D' in "D x y" mean the star at(x,y) is dim.When get a query as "Q X1 X2 Y1 Y2",you should tell Yifenfei how many bright stars there are in the region correspond X1,X2,Y1,Y2. There is only one case.   Input The first line contain a M(M <= 100000), then M line followed. each line start with a operational character. if the character is B or D,then two integer X,Y (0 <=X,Y<= 1000)followed. if the character is Q then four integer X1,X2,Y1,Y2(0 <=X1,X2,Y1,Y2<= 1000) followed.   Output For each query,output the number of bright stars in one line.   Sample Input 5 B 581 145 B 581 145 Q 0 600 0 200 D 581 145 Q 0 600 0 200   Sample Output 1 0    题目大意：   二维空间里，初始时一片空白，有如下几种操作：   1.B x y  ：将坐标为（x，y）的地方点亮   2.D x y  ：将坐标为（x，y）的地方熄灭   3. Q x x1 y y1 ：查询该矩形区域内点亮的星星个数  题解：树状数组模板  唯一坑点：当点（x，y）处已经点亮，则不再点亮  熄灭操作雷同。 #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; #define maxn 1005 int a[maxn][maxn],b[maxn][maxn]; int lowbit(int x) { return x&-x; } void update(int x,int y,int z) { int i,j; for(i=x;i<=maxn;i+=lowbit(i)) for(j=y;j<=maxn;j+=lowbit(j)) a[i][j]+=z; } int query(int x,int y) { int sum=0,i,j; for(i=x;i>0;i-=lowbit(i)) for(j=y;j>0;j-=lowbit(j)) sum+=a[i][j]; return sum; } int main() { int n,i,x,y,xx,yy; scanf("%d",&n); for(i=1;i<=n;i++) { char c[5]; scanf("%s",c); if(c[0]=='B') { scanf("%d%d",&x,&y); x++;y++; if(b[x][y]) continue; update(x,y,1); b[x][y]=1; } else if(c[0]=='D') { scanf("%d%d",&x,&y); x++;y++; if(!b[x][y]) continue; update(x,y,-1); b[x][y]=0; } else { scanf("%d%d%d%d",&x,&xx,&y,&yy); x++;y++; xx++;yy++; if(x<xx) swap(x,xx); if(y<yy) swap(y,yy); int ans=query(x,y)-query(x,yy-1)-(query(xx-1,y)-query(xx-1,yy-1)); printf("%d\n",ans); } } }