【BZOJ】1452: [JSOI2009]Count 树状数组

Description

Input

Output

Sample Input



Sample Output

1
2

HINT

  题解：

　　二维的树状数组啊+一维的颜色状态，然后直接做就好……实际上比照一维的树状数组就是多了一个for循环，然后查询操作的时候就相当于查询某一矩阵的大小，树状数组起到一个类似前缀和的作用。

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 #define lowbit(x) x&(-x)
 6 using namespace std;
 7 const int MAXN = 301;
 8 int a[MAXN][MAXN], Tree[101][MAXN][MAXN];
 9 int n, m;
10 void Modify(int x, int y, int s, int tar)
11 {
12     int i, j;
13     for (i = x; i <= n; i += lowbit(i))
14         for (j = y; j <= m; j += lowbit(j))
15             Tree[s][i][j] += tar;
16 }
17 inline int Query(int x, int y, int s)
18 {
19     int i, j;
20     int sum = 0;
21     for (i = x; i; i -= lowbit(i))
22         for (j = y; j; j -= lowbit(j))
23             sum += Tree[s][i][j];
24     return sum;
25 }
26 int main(int argc, char *argv[])
27 {
28     int c, i, j, x, Q, op, y;
29     int x1, y1, x2, y2;
30     int ans;
31     scanf("%d%d", &n, &m);
32     for (i = 1; i <= n; i++)
33         for (j = 1; j <= m; j++)
34         {
35         scanf("%d", &a[i][j]);
36         Modify(i, j, a[i][j], 1);
37         }
38     scanf("%d", &Q);
39     while (Q--)
40     {
41         scanf("%d", &op);
42         if (op == 1)
43         {
44             scanf("%d%d%d", &x, &y, &c);
45             Modify(x, y, a[x][y], -1);
46             a[x][y] = c;
47             Modify(x, y, a[x][y], 1);
48         }
49         else if (op == 2)
50         {
51             scanf("%d%d%d%d%d", &x1, &x2, &y1, &y2, &c);
52             ans = 0;
53             ans = Query(x1 - 1, y1 - 1, c) + Query(x2, y2, c) - Query(x1 - 1, y2, c) - Query(x2 , y1 - 1, c);
54             printf("%d\n", ans);
55         }
56     }
57     return 0;
58 }

 

posted @ 2016-09-19 09:10  BeyondW  阅读(187)  评论(0)  编辑  收藏  举报