【树状数组+二维】杭电 hdu 1892 See you~

/* THE PROGRAM IS MADE BY PYY */
/*----------------------------------------------------------------------------//
    Copyright (c) 2012 panyanyany All rights reserved.

    URL   : http://acm.hdu.edu.cn/showproblem.php?pid=1892
    Name  : 1892 See you~

    Date  : Monday, April 16, 2012
    Time Stage : one hour

    Result:
5787399	2012-04-16 21:51:19	Accepted	1892
203MS	8212K	2645 B
C++	pyy

5787369	2012-04-16 21:48:37	Wrong Answer	1892
218MS	8212K	2493 B
C++	pyy

5787344	2012-04-16 21:46:17	Time Limit Exceeded	1892
3000MS	8184K	2438 B
C++	pyy

5787212	2012-04-16 21:34:53	Time Limit Exceeded	1892
3000MS	4184K	2306 B
C++	pyy

5787126	2012-04-16 21:28:33	Time Limit Exceeded	1892
3000MS	4184K	2280 B
C++	pyy


Test Data :

Review :
第一次做的时候是984MS，第二次，免去初始化，果断就203MS了……
刚开始的几次TLE，是因为没有开另一个数组来记录每个格子的状态，直接通过
getsum(x1,y1)-getsum(x1-1,y1-1)来求每个格子的状态，结果就悲剧了……

二维以上的数组，经常要考虑容斥的问题，我就经常忽略掉，而且下标从1开始也经常
忽略掉，所以悲剧也经常来照顾我……
//----------------------------------------------------------------------------*/

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <vector>

#include <algorithm>
#include <iostream>
#include <queue>

using namespace std ;

#define MEM(a, v)        memset (a, v, sizeof (a))    // a for address, v for value
#define max(x, y)        ((x) > (y) ? (x) : (y))
#define min(x, y)        ((x) < (y) ? (x) : (y))

#define INF     (0x3f3f3f3f)
#define MAXN 1010
#define LESN 10002

#define L(x)	((x)<<1)
#define R(x)	(((x)<<1)|1)

#define DB    /##/
typedef __int64	LL;

int		tcase, q;
int		treeArr[MAXN][MAXN], map[MAXN][MAXN];

inline int lowbit(int x)
{
	return x & (-x);
}

void add(int x, int y, int val)
{
	int i, j;
	for (i = x; i < MAXN; i += lowbit(i))
	{
		for (j = y; j < MAXN; j += lowbit(j))
		{
			treeArr[i][j] += val;
		}
	}
}

int getsum(int x, int y)
{
	int i, j;
	int sum = 0;
	for (i = x; i > 0; i -= lowbit(i))
	{
		for (j = y; j > 0; j -= lowbit(j))
		{
			sum += treeArr[i][j];
		}
	}
	return sum;
}

void swap(int &x, int &y)
{
	int tmp = x;
	x = y;
	y = tmp;
}

int main()
{
	int i, j, k, x1, y1, x2, y2, n1, s;
	char c;
	while (scanf("%d", &tcase) != EOF)
	{
		for (k = 1; k <= tcase; ++k)
		{
			MEM(treeArr, 0);	// 不用两个for循环然后add(i,j,1)了
			MEM(map, 0);	// 不用给map元素全部初始化为1也行

			scanf("%d", &q);
			printf("Case %d:\n", k);
			while (q--)
			{
				getchar();
				scanf("%c %d %d", &c, &x1, &y1);
				++x1; ++y1;	// 题目的坐标是从0开始的，树状数组要求从1开始，
				switch(c)
				{
				case 'S':
					scanf("%d %d", &x2, &y2);
					++x2; ++y2;

					// 果然，(x2,y2) 可能比 (x1,y1) 要小……
					if (x2 < x1)
						swap(x1, x2);
					if (y2 < y1)
						swap(y1, y2);

					printf("%d\n", (getsum(x2, y2)+getsum(x1-1, y1-1)-
						getsum(x2, y1-1)-getsum(x1-1, y2))+
						(x2-x1+1)*(y2-y1+1));	// 在这里将初始每个格子的值加上
					break;
				case 'A':
					scanf("%d", &n1);
					map[x1][y1] += n1;
					add(x1, y1, n1);
					break;
				case 'D':
					scanf("%d", &n1);
					s = map[x1][y1];
					if (s+1 < n1)
						n1 = s + 1;
					add(x1, y1, -n1);
					map[x1][y1] -= n1;
					break;
				case 'M':
					scanf("%d %d %d", &x2, &y2, &n1);
					++x2; ++y2;

					s = map[x1][y1];
					if (s+1 < n1)
						n1 = s + 1;
					map[x1][y1] -= n1;
					map[x2][y2] += n1;
					add(x1, y1, -n1);
					add(x2, y2, n1);
					break;
				}
			}
		}
	}
	return 0;
}