hdu-1892-二维BIT

See you~

Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 6520    Accepted Submission(s): 2040

Problem Description

Now I am leaving hust acm. In the past two and half years, I learned so many knowledge about Algorithm and Programming, and I met so many good friends. I want to say sorry to Mr, Yin, I must leave now ~~>.<~~. I am very sorry, we could not advanced to the World Finals last year. 
When coming into our training room, a lot of books are in my eyes. And every time the books are moving from one place to another one. Now give you the position of the books at the early of the day. And the moving information of the books the day, your work is to tell me how many books are stayed in some rectangles. 
To make the problem easier, we divide the room into different grids and a book can only stayed in one grid. The length and the width of the room are less than 1000. I can move one book from one position to another position, take away one book from a position or bring in one book and put it on one position. 

 

Input

In the first line of the input file there is an Integer T(1<=T<=10), which means the number of test cases in the input file. Then N test cases are followed. 
For each test case, in the first line there is an Integer Q(1<Q<=100,000), means the queries of the case. Then followed by Q queries. 
There are 4 kind of queries, sum, add, delete and move. 
For example: 
S x1 y1 x2 y2 means you should tell me the total books of the rectangle used (x1,y1)-(x2,y2) as the diagonal, including the two points. 
A x1 y1 n1 means I put n1 books on the position (x1,y1) 
D x1 y1 n1 means I move away n1 books on the position (x1,y1), if less than n1 books at that position, move away all of them. 
M x1 y1 x2 y2 n1 means you move n1 books from (x1,y1) to (x2,y2), if less than n1 books at that position, move away all of them. 
Make sure that at first, there is one book on every grid and 0<=x1,y1,x2,y2<=1000,1<=n1<=100. 

 

Output

At the beginning of each case, output "Case X:" where X is the index of the test case, then followed by the "S" queries. 
For each "S" query, just print out the total number of books in that area. 

 

Sample Input

2
3
S 1 1 1 1
A 1 1 2
S 1 1 1 1
3
S 1 1 1 1
A 1 1 2
S 1 1 1 2

 

Sample Output

Case 1:
1
3
Case 2:
1
4

 

Author

Sempr|CrazyBird|hust07p43

 

Source

HDU 2008-4 Programming Contest

　　

　　二维BIT的模板题，注意S时给出的两个点要自己转换一下保证一个是左上角一个是右下角。

　二维的BIT里面的每个节点是一个矩阵，宽度就是x的lowbit，长度就是y的lowbit，右下角就是(X,Y)。

　　

 #include<bits/stdc++.h>
 using namespace std;
 #define ULL unsigned long long
 #define LL long long
 LL C[][];
 int A[][];
 inline int lowbit(int x){return x&-x;}
 void add(int x,int y,int d){
     for(int i=x;i<=;i+=lowbit(i))
         for(int j=y;j<=;j+=lowbit(j))
             C[i][j]+=d;
 }
 LL sum(int x,int y){
     LL r=;
     for(int i=x;i>;i-=lowbit(i))
         for(int j=y;j>;j-=lowbit(j))
             r+=C[i][j];
     return r;
 }
 int main(){
     int t,x1,x2,y1,y2,n,q,cas=;
     char ch[];
     scanf("%d",&t);
     while(t--){
         memset(C,,sizeof(C));
         for(int i=;i<=;++i)
             for(int j=;j<=;++j) add(i,j,),A[i][j]=;
         scanf("%d",&q);
         printf("Case %d:\n",++cas);
         while(q--){
             scanf("%s",ch);
             if(ch[]=='S'){
                 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                 x1++,y1++,x2++,y2++;
                 if(x1>x2) swap(x1,x2);
                 if(y1>y2) swap(y1,y2);
                 printf("%lld\n",sum(x2,y2)-sum(x1-,y2)-sum(x2,y1-)+sum(x1-,y1-));
             }
             else if(ch[]=='A'){
                 scanf("%d%d%d",&x1,&y1,&n);
                 x1++,y1++;
                 add(x1,y1,n);
                 A[x1][y1]+=n;
             }
             else if(ch[]=='D'){
                 scanf("%d%d%d",&x1,&y1,&n);
                 x1++,y1++;
                 n=min(n,A[x1][y1]);
                 add(x1,y1,-n);
                 A[x1][y1]-=n;
             }
             else if(ch[]=='M'){
                 scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&n);
                 x1++,y1++,x2++,y2++;
                 n=min(n,A[x1][y1]);
                 add(x1,y1,-n),A[x1][y1]-=n;
                 add(x2,y2,n),A[x2][y2]+=n;
             }
         }
     }
     return ;
 }