POJ 2155（二维树状数组）

Language: Default Matrix Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 29131   Accepted: 10630 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 题意：给你一个n*n的矩阵，初始值为0，现在有两种操作，’C‘ 翻转给定矩阵的元素（0变为1,1变为0），’Q‘ 查询某个点的值。 分析：难点是修改矩阵内的值，其实只要修改(x1,y1)，(x1,y2+1),(x2+1,y2+1),(x2+1,y1);这四点值即可。 对于某一块来说，假如翻转了偶数次，就相当与没有改变，所以只需要知道某点翻转的总次数sum，答案就是sum%2了。树状数组里 得出的其实就是翻转总次数。 可以结合这位大佬的图来理解：http://blog.csdn.net/zxy_snow/article/details/6264135 代码： #include #include #include #include using namespace std; typedef long long LL; const int maxn=1005; int tree[maxn][maxn]; int n; int lowbit(int x) {     return x&-x; } void add(int pos1,int pos2) {     for(int i=pos1;i<=n;i+=lowbit(i))     {         for(int j=pos2;j<=n;j+=lowbit(j))         {             tree[i][j]++;         }     } } int GetSum(int pos1,int pos2) {     int ans=0;     for(int i=pos1;i>0;i-=lowbit(i))     {         for(int j=pos2;j>0;j-=lowbit(j))         {             ans+=tree[i][j];         }     }     return ans; } int main() {     int T;     scanf("%d",&T);     while(T--)     {         memset(tree,0,sizeof(tree));         int q;         scanf("%d%d",&n,&q);         while(q--)         {             char s[10];             scanf("%s",s);             if(s[0]=='C')             {                 int x1,y1,x2,y2;                 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);                 add(x1,y1);                 add(x1,y2+1);                 add(x2+1,y2+1);                 add(x2+1,y1);             }             if(s[0]=='Q')             {                 int x,y;                 scanf("%d%d",&x,&y);                 printf("%d\n",GetSum(x,y)%2);             }         }         printf("\n");     }     return 0; }