See you~ （hdu1892 二维树状数组模板）

See you~

Time Limit : 5000/3000ms (Java/Other)   Memory Limit : 65535/32768K (Java/Other)

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：有一个书架（空间是二维的），初始时每个位置（坐标）上都有1本书，现在有4种操作：“S”：输入x1,y1,x2,y2，求以（x1,y1）,（x2,y2）为对角线顶点的矩形区域内共有几本书；“A”：输入x,y,n，往（x,y）这个位置加进去n本书；“D”：输入x,y,n，在（x,y）这个位置拿掉n本书；“M”：输入x1,y1,x2,y2,n，从（x1,y1）这个位置拿n本书放到（x2,y2）这个位置；

//思路：其实就是一个简单的二维树状数组模板。“S”就是求和，“A”、“D”、“M”其实都可以看成添加操作（减的话就是加个负的）。主要要注意的就是些细节，见代码。

//二维树状数组模板题#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>using namespace std;#define size 1010int n;//这里的+1很有必要，不然会出现各种奇怪的错误...int c[size + 1][size + 1];int lowbit(int x){return x&(-x);}void add(int x, int y, int val){for (int i = x; i <= size; i += lowbit(i)){for (int j = y; j <= size; j += lowbit(j)){c[i][j] += val;}}}int sum(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 += c[i][j];}}return sum;}void Init(){memset(c, 0, sizeof(c));for (int i = 1; i <= size; i++){for (int j = 1; j <= size; j++){add(i, j, 1);}}}int main(){int i;int T, cas = 1;cin >> T;while (T--){char a;int x1, y1, x2, y2;int num;Init();scanf("%d", &n);printf("Case %d:\n", cas++);getchar();for (i = 1; i <= n; i++){//用scanf %c 的话记得getchar()scanf("%c", &a);//下面坐标++是因为题目里说x,y是可以取到0的//但树状数组中下标不能为0，所以++//x1,y1 -1是考虑顶点，题目要求包含顶点if (a == '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);int temp = sum(x2, y2) + sum(x1 - 1, y1 - 1) - sum(x2, y1 - 1) - sum(x1 - 1, y2);printf("%d\n", temp);}else if (a == 'A'){scanf("%d%d%d", &x1, &y1, &num);x1++;y1++;add(x1, y1, num);}else if (a == 'D'){scanf("%d%d%d", &x1, &y1, &num);x1++;y1++;int temp = sum(x1, y1) + sum(x1 - 1, y1 - 1) - sum(x1 - 1, y1) - sum(x1, y1 - 1);if (num > temp)num = temp;add(x1, y1, -num);}else if (a == 'M'){scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &num);x1++;y1++;x2++;y2++;int temp = sum(x1, y1) + sum(x1 - 1, y1 - 1) - sum(x1 - 1, y1) - sum(x1, y1 - 1);if (num > temp)num = temp;add(x1, y1, -num);add(x2, y2, num);}getchar();}}return 0;}