hdu 5465 树状数组

因为xor具有前缀和性质，所以可以建立一颗二维树状数组，写起来只是稍微有点改变，感觉此题还是很不错的，结合了数据结构和博弈。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int N = 501;
int a[N][N];
int c[N][N];
int n, m, q;

int lb( int i )
{
    return i & -i;
}

void update( int i, int j, int val )
{
    for ( int ii = i; ii <= n; ii += lb(ii) )
    {
        for ( int jj = j; jj <= m; jj += lb(jj) )
        {
            c[ii][jj] ^= val;
        }
    }
}

int sum( int i, int j )
{
    int res = 0;
    for ( int ii = i; ii > 0; ii -= lb(ii) )
    {
        for ( int jj = j; jj > 0; jj -= lb(jj) )
        {
            res ^= c[ii][jj];
        }
    }
    return res;
}

int query( int x1, int y1, int x2, int y2 )
{
    int p = sum( x2, y2 );
    int q = sum( x1 - 1, y2 );
    int r = sum( x2, y1 - 1 );
    int s = sum( x1 - 1, y1 - 1 );
    return p ^ q ^ r ^ s;
}

int main ()
{
    int t;
    scanf("%d", &t);
    while ( t-- )
    {
        scanf("%d%d%d", &n, &m, &q);
        memset( a, 0, sizeof(a) );
        memset( c, 0, sizeof(c) );
        for ( int i = 1; i <= n; i++ )
        {
            for ( int j = 1; j <= m; j++ )
            {
                scanf("%d", &a[i][j]);
                update( i, j, a[i][j] );
            }
        }
        while ( q-- )
        {
            int opt;
            scanf("%d", &opt);
            if ( opt == 1 )
            {
                int x1, y1, x2, y2;
                scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
                if ( query( x1, y1, x2, y2 ) )
                {
                    puts("Yes");
                }
                else
                {
                    puts("No");
                }
            }
            else
            {
                int x, y, z;
                scanf("%d%d%d", &x, &y, &z);
                update( x, y, a[x][y] ^ z );
                a[x][y] = z;
            }
        }
    }
    return 0;
}