poj1195--Mobile phones(二维树状数组)

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Description

Input

The input is read from standard input as integers and the answers to the queries are written to standard output as integers. The input is encoded as follows. Each input comes on a separate line, and consists of one instruction integer and a number of parameter integers according to the following table. 

The values will always be in range, so there is no need to check them. In particular, if A is negative, it can be assumed that it will not reduce the square value below zero. The indexing starts at 0, e.g. for a table of size 4 * 4, we have 0 <= X <= 3 and 0 <= Y <= 3. 

Table size: 1 * 1 <= S * S <= 1024 * 1024 
Cell value V at any time: 0 <= V <= 32767 
Update amount: -32768 <= A <= 32767 
No of instructions in input: 3 <= U <= 60002 
Maximum number of phones in the whole table: M= 2^30 

Output

Sample Input

0 4
1 1 2 3
2 0 0 2 2 
1 1 1 2
1 1 2 -1
2 1 1 2 3 
3

Sample Output

3
4

Source

简单的二维树状数组求区域和，注意xy的坐标，要保证从1开始。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxn 1100
int c[maxn][maxn] , n ;
int f(int x)
{
    return x & -x ;
}
void add(int x,int y,int a)
{
    int i , j ;
    for(i = x ; i <= n ; i += f(i))
    {
        for(j = y ; j <= n ; j += f(j) )
            c[i][j] += a ;
    }
    return ;
}
int sum(int x,int y)
{
    int ans = 0 , i , j ;
    for(i = x ; i >= 1 ; i -= f(i) )
    {
        for(j = y ; j >= 1 ; j -= f(j) )
            ans += c[i][j];
    }
    return ans ;
}
int main()
{
    int k , i , j , x1 , y1 , x2 , y2 , a , temp ;
    scanf("%d %d", &k, &n);
    memset(c,0,sizeof(c));
    while(1)
    {
        scanf("%d", &k);
        if(k == 3)
            break;
        if(k == 1)
        {
            scanf("%d %d %d", &x1, &y1, &a);
            x1++ ;
            y1++ ;
            add(x1,y1,a);
            continue ;
        }
        scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
        x1++ ;
        y1++ ;
        x2++ ;
        y2++ ;
        temp = sum(x2,y2) + sum(x1-1,y1-1) - sum(x1-1,y2) - sum(x2,y1-1);
        printf("%d\n", temp);
    }
    return 0;
}

poj1195--Mobile phones(二维树状数组)

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原文地址：http://blog.csdn.net/winddreams/article/details/39504837