Matrix
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 14123 | Accepted: 5337 |
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
12 10C 2 1 2 2Q 2 2C 2 1 2 1Q 1 1C 1 1 2 1C 1 2 1 2C 1 1 2 2Q 1 1C 1 1 2 1Q 2 1
Sample Output
1001
Source
,Lou Tiancheng
此题大意：给你一个2维平面，C:(X1,Y1)(X2,Y2)就是将这里范围的点改变：0->1、1->0，Q:(x,y)就是问这个点是什么。
#include#include #include using namespace std;const int N=1010;int n,arr[N][N];int lowbit(int x){ return x&(-x);}void update(int i,int j,int val){ while(i<=n){ int tmpj=j; while(tmpj<=n){ arr[i][tmpj]+=val; tmpj+=lowbit(tmpj); } i+=lowbit(i); }}int Sum(int i,int j){ int ans=0; while(i>0){ int tmpj=j; while(tmpj>0){ ans+=arr[i][tmpj]; tmpj-=lowbit(tmpj); } i-=lowbit(i); } return ans;}int main(){ //freopen("input.txt","r",stdin); int t,q; scanf("%d",&t); char op[3]; while(t--){ scanf("%d%d",&n,&q); memset(arr,0,sizeof(arr)); int x1,y1,x2,y2; while(q--){ scanf("%s",op); if(op[0]=='C'){ scanf("%d%d%d%d",&x1,&y1,&x2,&y2); update(x2+1,y2+1,1); update(x2+1,y1,1); update(x1,y2+1,1); update(x1,y1,1); }else if(op[0]=='Q'){ scanf("%d%d",&x1,&y1); printf("%d\n",Sum(x1,y1)&1); //该点的值就是sum(x,y) } } if(t!=0) printf("\n"); } return 0;}
#include#include #include using namespace std;const int N=1010;int array[N][N];int n;int lowbit(int x){ return x&(-x);}void update(int i,int j){ int tmp; while(i>0){ tmp=j; while(tmp>0){ array[i][tmp]^=1; tmp-=lowbit(tmp); } i-=lowbit(i); }}int query(int i,int j){ int ans=0; int tmp; while(i<=n){ tmp=j; while(tmp<=n){ ans^=array[i][tmp]; tmp+=lowbit(tmp); } i+=lowbit(i); } return ans;}int main(){ //freopen("input.txt","r",stdin); int t,m; char op; scanf("%d",&t); while(t--){ memset(array,0,sizeof(array)); scanf("%d%d",&n,&m); int x1,y1,x2,y2; while(m--){ getchar(); scanf("%c",&op); if(op=='C'){ scanf("%d%d%d%d",&x1,&y1,&x2,&y2); update(x2,y2); update(x1-1,y2); update(x2,y1-1); update(x1-1,y1-1); }else{ scanf("%d%d",&x1,&y1); printf("%d\n",query(x1,y1)); } } if(t!=0) printf("\n"); } return 0;}