Stars Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/65536 K (Java/Others)
Total Submission(s): 2428 Accepted Submission(s): 1006
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 Author teddy Source 百万秦关终属楚 题意：就是问在给定区域内，有多少点亮着的星星 解题思路：二维树状数组 实时更新和求和 利用矩形的计算方法，求出某个区域的值 ac代码： #include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 1009
using namespace std;
int sum[maxn][maxn];
int lowbit(int x)
{
return x&-x;
}
int sm(int x,int y)
{
int ans=0;
for(int i=x;i>0;i-=lowbit(i))
for(int j=y;j>0;j-=lowbit(j))
{
ans+=sum[i][j];
}
return ans;
}
void updata(int x,int y,int val)
{
for(int i=x;i<=maxn;i+=lowbit(i))
for(int j=y;j<=maxn;j+=lowbit(j))
{
sum[i][j]+=val;
}
}
int get_val(int i,int j)
{
int ans= sm(i,j)-sm(i-1,j)-sm(i,j-1)+sm(i-1,j-1);
return ans;
}
int main()
{
int m;
memset(sum,0,sizeof(sum));
char s[5];
int ans=0;
cin>>m;
int x1,y1,x2,y2;
while(m--)
{
scanf("%s",s);
if(s[0]=='B')
{
scanf("%d%d",&x1,&y1);
x1++;
y1++;
if(get_val(x1,y1)==0)
updata(x1,y1,1);
}
if(s[0]=='D')
{
scanf("%d%d",&x1,&y1);
x1++;
y1++;
if(get_val(x1,y1)==1)
updata(x1,y1,-1);
}
if(s[0]=='Q')
{
scanf("%d%d%d%d",&x1,&x2,&y1,&y2);
if(x1>x2) swap(x1,x2);
if(y1>y2) swap(y1,y2);
x1++,y1++,x2++,y2++;
ans=sm(x2,y2)-sm(x1-1,y2)-sm(x2,y1-1)+sm(x1-1,y1-1);
printf("%d\n",ans);
}
}
return 0;
}
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