POJ 1195 Mobile phones (树状数组)

题意：给定n*n矩阵，和几种在线操作，包括对某一点(x,y)值修改，查询一个矩形(x1,y1,x2,y2)的元素和。

思路：典型的在线查询，可用树状数组实现，查询矩形和时，稍微注意以下就可以了:

　　sum(x2,y2)+sum(x1-1,y1-1)-sum(x1-1,y2)-sum(x2,y1-1);

还要注意树状数组的修改操作modify(index,delta)中的index要>0。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <memory.h>
#include <cmath>
#include <bitset>
#include <queue>
#include <vector>
using namespace std;

const int BORDER = (1<<20)-1;
const int MAXSIZE = 37;
const int MAXN = 1100;
const int INF = 1000000000;
#define CLR(x,y) memset(x,y,sizeof(x))
#define ADD(x) x=((x+1)&BORDER)
#define IN(x) scanf("%d",&x)
#define OUT(x) printf("%d\n",x)
#define MIN(m,v) (m)<(v)?(m):(v)
#define MAX(m,v) (m)>(v)?(m):(v)
#define ABS(x) ((x)>0?(x):-(x))

int tre[MAXN][MAXN];
int n,n_tre,t_case;

int lowbit(int x)
{
    return x&(-x);
}
void modify(const int& x,const int& y,const int delta)
{
    for(int i = x; i <= n_tre; i += lowbit(i))
        for(int j = y; j <= n_tre; j += lowbit(j))
        {
            tre[i][j] += delta;
        }
}
int get_sum(const int& x,const int& y)
{
    int sum = 0;
    for(int i = x; i > 0; i -= lowbit(i))
        for(int j = y; j > 0; j -= lowbit(j))
            sum += tre[i][j];
    return sum;
}
int init()
{
    CLR(tre,0);
    return 0;
}
int work()
{
    int i,j,x1,x2,y1,y2,x,y,val,tmp,ans;
    char s[3];
    n_tre = n+5;
    while(scanf("%s",s))
    {
        if(s[0] == '3')
            break;
        if(s[0] == '1')
        {
            scanf("%d%d%d",&x,&y,&val);
            modify(x+1,y+1,val);
        }else 
        {
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            ans = get_sum(x2+1,y2+1)+get_sum(x1,y1)-
                get_sum(x1,y2+1)-get_sum(x2+1,y1);
            OUT(ans);
        }
    }
    return 0;
}
int main()
{
    while(scanf("%d %d",&t_case,&n)!=EOF)
    {
        init();
        work();
    }
    return 0;
}