# 【POJ2155】【二维树状数组】Matrix

Description

Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1 <= i, j <= N).

We can change the matrix in the following way. Given a rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2), we change all the elements in the rectangle by using "not" operation (if it is a '0' then change it into '1' otherwise change it into '0'). To maintain the information of the matrix, you are asked to write a program to receive and execute two kinds of instructions.

1. C x1 y1 x2 y2 (1 <= x1 <= x2 <= n, 1 <= y1 <= y2 <= n) changes the matrix by using the rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2).
2. Q x y (1 <= x, y <= n) querys A[x, y].

Input

The first line of the input is an integer X (X <= 10) representing the number of test cases. The following X blocks each represents a test case.

The first line of each block contains two numbers N and T (2 <= N <= 1000, 1 <= T <= 50000) representing the size of the matrix and the number of the instructions. The following T lines each represents an instruction having the format "Q x y" or "C x1 y1 x2 y2", which has been described above.

Output

For each querying output one line, which has an integer representing A[x, y].

There is a blank line between every two continuous test cases.

Sample Input

1
2 10
C 2 1 2 2
Q 2 2
C 2 1 2 1
Q 1 1
C 1 1 2 1
C 1 2 1 2
C 1 1 2 2
Q 1 1
C 1 1 2 1
Q 2 1


Sample Output

1
0
0
1


Source

POJ Monthly,Lou Tiancheng
【分析】

 1 #include <iostream>
2 #include <cstdio>
3 #include <algorithm>
4 #include <cstring>
5 #include <vector>
6 #include <utility>
7 #include <iomanip>
8 #include <string>
9 #include <cmath>
10 #include <map>
11
12 const int MAXN = 1000 + 10;
13 const int MAX = 32000 + 10;
14 using namespace std;
15 int n, m;//m为操作次数
16 int C[MAXN][MAXN];
17
18 int lowbit(int x){return x&-x;}
19 /*int sum(int x, int y){
20     int cnt = 0, tmp;
21     while (x > 0){
22           tmp = y;
23           while (tmp > 0){
24                 cnt += C[x][tmp];
25                 tmp -= lowbit(tmp);
26           }
27           x -= lowbit(x);
28     }
29     return cnt;
30 }
31 void add(int x, int y, int val){
32      int tmp;
33      while (x <= 1000){
34            tmp = y;
35            while (tmp <= 1000){
36                  C[x][tmp] += val;
37                  tmp += lowbit(tmp);
38            }
39            x += lowbit(x);
40      }
41      return;
42 }*/
43 void add(int x,int y)  {
44     int i,k;
45     for(i=x; i<=n; i+=lowbit(i))
46         for(k=y; k<=n; k+=lowbit(k))
47             C[i][k]++;
48 }
49 int sum(int x,int y)  {
50     int i,k,cnt = 0;
51     for(i=x; i>0; i-=lowbit(i))
52         for(k=y; k>0; k-=lowbit(k))
53             cnt += C[i][k];
54     return cnt;
55 }
56
57 void work(){
58      scanf("%d%d", &n, &m);
59      for (int i = 1; i <= m; i++){
60          char str[2];
61          scanf("%s", str);
62          if (str[0] == 'Q'){
63             int x, y;
64             scanf("%d%d", &x, &y);
65             //x++;y++;
66             printf("%d\n", sum(x, y)%2);
67          }else if (str[0] == 'C'){
68                int x1, y1, x2, y2;
69                scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
70                x1++;y1++;x2++;y2++;
74                add(x1 - 1, y1 - 1);
75          }
76      }
77 }
78
79 int main(){
80     int T;
81     #ifdef LOCAL
82     freopen("data.txt",  "r",  stdin);
83     freopen("out.txt",  "w",  stdout);
84     #endif
85     scanf("%d", &T);
86     while (T--){
87           memset(C, 0, sizeof(C));
88           work();
89           printf("\n");
90     }
91     return 0;
92 }
View Code

posted @ 2015-03-07 11:42  TCtower  阅读(200)  评论(0编辑  收藏  举报