POJ-1195-Mobile phones-裸二维树状数组（单点更新,矩阵求和）

http://poj.org/problem?id=1195

题意就是 对单点更新，每次对一个子矩阵求和啦。。。。

仅要实现这个功能的话，二维树状数组比二维线段树代码短了不是一点半点......

一维树状数组的更新

void add(int x,int val)
{
    for(;x<=n;x+=lowbit(x))
    {
        num[x]+=val;
    }
}

二维的更新

void add(int x,int y,int a)
	{
		int i,j;
		for (i=x;i<=s;i+=lowbit(i))
		{
			for (j=y;j<=s;j+=lowbit(j))
				num[i][j]+=a;
		}
	}

一维的查询

int query(int x)
{
    int ans=0;
    for(;x>0;x-=lowbit(x))
    {
        ans+=a[i];
    }
    return ans;
}

二维的查询

	int query(int x,int y)
	{
		int ans=0,i,j;
		for (i=x;i>0;i-=lowbit(i))
		{
			for (j=y;j>0;j-=lowbit(j))
				ans+=num[i][j];
		}
		return ans;
	}  

注意下标从0开始

#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <set>
#include <map> 
#include <vector>
#include <iostream>
using namespace std; 
#define ll int
const double pi=acos(-1.0);
double eps=1e-5;    
int max(int a,int b)
{return a>b?a:b;}
int min(int a,int b)
{return a<b?a:b;}
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
int n,m;
const int maxn=1000+50; 
int s; 
struct tree
{
	int num[maxn][maxn];
	inline	int lowbit(int x)
	{return x&-x;}
	void add(int x,int y,int a)
	{
		int i,j;
		for (i=x;i<=s;i+=lowbit(i))
		{
			for (j=y;j<=s;j+=lowbit(j))
				num[i][j]+=a;
		}
	}
	int query(int x,int y)
	{
		int ans=0,i,j;
		for (i=x;i>0;i-=lowbit(i))
		{
			for (j=y;j>0;j-=lowbit(j))
				ans+=num[i][j];
		}
		return ans;
	}  
};


tree tp;
int main()
{  
	int i;
	int op; 
	int a,x,y;
	int l,b,r,t;
	while(scanf("%d",&op)!=EOF)
	{
		if (op==3) break;
		if (op==0)
		{
			scanf("%d",&s);
			memset(tp.num,0,sizeof(tp.num));
		}
		else
			if (op==1)
			{
				scanf("%d%d%d",&x,&y,&a);
				x++,y++;
				tp.add(x,y,a);
			}
			else
			{
				scanf("%d%d%d%d",&l,&b,&r,&t);
				l++,b++,r++,t++;
				int ret=tp.query(r,t);
				ret-=tp.query(l-1,t);
				ret-=tp.query(r,b-1);
				ret+=tp.query(l-1,b-1);
				printf("%d\n",ret);
			} 
			
	}
		  return 0; 
}