poj2155二维线段树

Matrix

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 27530 Accepted: 10050

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

100
1
AC代码：
#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1#define N 1010using namespace std;int t,n,q,ans;char s[5];int sum[N<<2][N<<2];void update2(int L,int R,int l,int r,int rtx,int rty){        if(L<=l&&r<=R)        {                sum[rtx][rty]^=1;                return ;        }        int mid=(l+r)>>1;        if(L>mid)                update2(L,R,mid+1,r,rtx,rty<<1|1);        else if(R<=mid)                update2(L,R,l,mid,rtx,rty<<1);        else        {                update2(L,mid,l,mid,rtx,rty<<1);                update2(mid+1,R,mid+1,r,rtx,rty<<1|1);        }}void update(int x1,int x2,int y1,int y2,int l,int r,int rt){        if(x1<=l&&r<=x2)        {                update2(y1,y2,1,n,rt,1);                return ;        }        int mid=(l+r)>>1;        if(x1>mid)                update(x1,x2,y1,y2,rson);        else if(x2<=mid)                update(x1,x2,y1,y2,lson);        else        {                update(x1,mid,y1,y2,lson);                update(mid+1,x2,y1,y2,rson);        }}void query2(int y1,int l,int r,int rtx,int rty){        ans^=sum[rtx][rty];        if(l==r)        {                return;        }        int mid=(l+r)>>1;        if(y1<=mid)                query2(y1,l,mid,rtx,rty<<1);        else                query2(y1,mid+1,r,rtx,rty<<1|1);}void query(int x1,int y1,int l,int r,int rt){        query2(y1,1,n,rt,1);        if(l==r)        {                return;        }        int mid=(l+r)>>1;        if(x1<=mid)                query(x1,y1,l,mid,rt<<1);        else                query(x1,y1,mid+1,r,rt<<1|1);}int main(){        scanf("%d",&t);        while(t--)        {                memset(sum,0,sizeof(sum));                scanf("%d%d",&n,&q);                for(int i=1;i<=q;i++)                {                        scanf("%s",s);                        if(s[0]=='C')                        {                                int x1,y1,x2,y2;                                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);                                update(x1,x2,y1,y2,1,n,1);                        }                        else                        {                                ans=0;                                int x1,y1;                                scanf("%d%d",&x1,&y1);                                query(x1,y1,1,n,1);                                printf("%d\n",ans);                        }                }                printf("\n");        }}