hdu 1892 See you~(二维树状数组）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1892点击打开链接

See you~

Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 5296    Accepted Submission(s): 1688

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

23S 1 1 1 1A 1 1 2S 1 1 1 13S 1 1 1 1A 1 1 2S 1 1 1 2

 

Sample Output

Case 1:13Case 2:14

题意：在二维空间里搬书 然后需要你写删除 计算区间和 增加 移动 的操作 其中 删除可以通过增加该位置的负数实现  移动既是删除和增加的总和 

这题坑点不少 其中0位置也要计算因此只能将方格数+1 防止在lowbit里死循环  另外可以另开一个二维数组记忆化每个点的个数 因为删除数据如果要删除的比本来的数大 只能到0 因此记忆化可以比较快速判断 这里不想用记忆化 就得多次计算getsum值 另外因为getsum是（0，0）~（x，y）的值 所以得进行计算 具体计算矩形的面积加减不详述了

#include <stdio.h>#include <stdlib.h>#include <iostream>#include<algorithm>#include <math.h>#include <string.h>#include <limits.h>#include <string>#include <queue>#include <stack>#include <set>#include <vector>using namespace std;int tree[1111][1111];int lowbit (int x){    return x&-x;}int getsum(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;}void add(int x,int y,int num){    for(int i=x;i<=1010;i+=lowbit(i))        for(int j=y;j<=1010;j+=lowbit(j))            tree[i][j]+=num;}int main(){    int t =0;    cin >> t ;    for(int iii=1;iii<=t;iii++)    {        memset(tree,0,sizeof(tree));        printf("Case %d:\n",iii);        for(int ii=0;ii<=1000;ii++)            for(int j=0;j<=1000;j++)            {                add(ii+1,j+1,1);            }        int tt;        cin >> tt ;        while(tt--)        {            char c;            cin >> c;            if(c=='S')            {                int x1,x2,y1,y2;                cin >> x1 >> y1 >> x2 >> y2;                int xx1,xx2,yy1,yy2;                xx1=min(x1,x2);                xx2=max(x1,x2);                yy1=min(y1,y2);                yy2=max(y1,y2);                printf("%d\n",getsum(xx2+1,yy2+1)-getsum(xx2+1,yy1)-getsum(xx1,yy2+1)+getsum(xx1,yy1));            }            else if(c=='A')            {                int x,y,num;                cin >> x >> y >> num;                add(x+1,y+1,num);            }            else if(c=='M')            {                int x1,x2,y1,y2,num;                cin >> x1 >> y1 >> x2 >> y2 >> num;                if((getsum(x1+1,y1+1)-getsum(x1,y1+1)-getsum(x1+1,y1)+getsum(x1,y1)<num))                    num=(getsum(x1+1,y1+1)-getsum(x1,y1+1)-getsum(x1+1,y1)+getsum(x1,y1));                add(x1+1,y1+1,-num);                add(x2+1,y2+1,num);            }            else if(c=='D')            {                int x,y,num;                cin >> x >> y >> num;                if((getsum(x+1,y+1)-getsum(x,y+1)-getsum(x+1,y)+getsum(x,y))>num)                    add(x+1,y+1,-num);                else                    add(x+1,y+1,-(getsum(x+1,y+1)-getsum(x,y+1)-getsum(x+1,y)+getsum(x,y)));            }        }    }}