Stars

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/65536 K (Java/Others) Total Submission(s): 975    Accepted Submission(s): 420

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

百萬秦關終屬楚

樹狀陣列，用到二維，當時總的的來說，還算簡單。。。

題目要求求矩形裡星星中的個數//

給定某兩個座標對角座標，球該矩形的星星個數..

對於x1,x2,y1,y2.。我們不知道其大小，所以需要進行比較

得到大小之後，我們就可以求該巨型的了，像下面的圖一樣..

由此貼出程式碼吧：

 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 #include<string.h>
 4 #define maxn  1005
 5 #define lowbit(x) ((x)&(-x))
 6 int aa[maxn][maxn];
 7 bool bb[maxn][maxn];
 8 
 9 void ope(int x,int y,int val)
10 {
11     int j;
12     if(val==1)
13     {
14         if(bb[x][y])    return ;
15         bb[x][y]=true;
16     }
17     else
18     {
19       if(bb[x][y]==false)
20            return ;
21       bb[x][y]=false;
22     }
23     while(x<maxn){
24        j=y;
25      while(j<maxn){
26        aa[x][j]+=val;
27        j+=lowbit(j);
28      }
29       x+=lowbit(x);
30     }
31 }
32 int sum(int x,int y)
33 {
34   int ans=0 ,j;
35   while(x>0){
36      j=y;
37     while(j>0){
38      ans+=aa[x][j];
39      j-=lowbit(j);
40     }
41      x-=lowbit(x);
42   }
43    return ans;
44 }
45 struct node
46 {
47     int x;
48     int y;
49 };
50 int main()
51 {
52     int test,res;
53     char str[2];
54     node a,b;
55     memset(aa,0,sizeof(aa));
56     memset(bb,0,sizeof(bb));
57     scanf("%d",&test);
58     while(test--)
59     {
60         scanf("%s",str);
61         if(str[0]=='Q')
62         {
63             scanf("%d%d%d%d",&a.x,&b.x,&a.y,&b.y);
64             if(a.x>b.x){
65                 a.x^=b.x;
66                 b.x^=a.x;
67                 a.x^=b.x;
68             }
69             if(a.y>b.y){
70                 a.y^=b.y;
71                 b.y^=a.y;
72                 a.y^=b.y;
73             }
74             b.x++;
75             b.y++;
76             res=sum(b.x,b.y)-sum(a.x,b.y)+sum(a.x,a.y)-sum(b.x,a.y);
77             printf("%dn",res);
78         }
79         else
80         {
81           scanf("%d%d",&a.x,&a.y);
82           a.x++;  //ÓÒÒÆһλ
83           a.y++;
84           if(str[0]=='B')
85              ope(a.x,a.y,1);
86           else
87              ope(a.x,a.y,-1);
88         }
89     }
90     return 0;
91 }