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。

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 using namespace std;
 5
 6 const int N = 501;
 7 int a[N][N];
 8 int c[N][N];
 9 int n, m, q;
10
11 int lb( int i )
12 {
13     return i & -i;
14 }
15
16 void update( int i, int j, int val )
17 {
18     for ( int ii = i; ii <= n; ii += lb(ii) )
19     {
20         for ( int jj = j; jj <= m; jj += lb(jj) )
21         {
22             c[ii][jj] ^= val;
23         }
24     }
25 }
26
27 int sum( int i, int j )
28 {
29     int res = 0;
30     for ( int ii = i; ii > 0; ii -= lb(ii) )
31     {
32         for ( int jj = j; jj > 0; jj -= lb(jj) )
33         {
34             res ^= c[ii][jj];
35         }
36     }
37     return res;
38 }
39
40 int query( int x1, int y1, int x2, int y2 )
41 {
42     int p = sum( x2, y2 );
43     int q = sum( x1 - 1, y2 );
44     int r = sum( x2, y1 - 1 );
45     int s = sum( x1 - 1, y1 - 1 );
46     return p ^ q ^ r ^ s;
47 }
48
49 int main ()
50 {
51     int t;
52     scanf("%d", &t);
53     while ( t-- )
54     {
55         scanf("%d%d%d", &n, &m, &q);
56         memset( a, 0, sizeof(a) );
57         memset( c, 0, sizeof(c) );
58         for ( int i = 1; i <= n; i++ )
59         {
60             for ( int j = 1; j <= m; j++ )
61             {
62                 scanf("%d", &a[i][j]);
63                 update( i, j, a[i][j] );
64             }
65         }
66         while ( q-- )
67         {
68             int opt;
69             scanf("%d", &opt);
70             if ( opt == 1 )
71             {
72                 int x1, y1, x2, y2;
73                 scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
74                 if ( query( x1, y1, x2, y2 ) )
75                 {
76                     puts("Yes");
77                 }
78                 else
79                 {
80                     puts("No");
81                 }
82             }
83             else
84             {
85                 int x, y, z;
86                 scanf("%d%d%d", &x, &y, &z);
87                 update( x, y, a[x][y] ^ z );
88                 a[x][y] = z;
89             }
90         }
91     }
92     return 0;
93 }