POJ

Matrix

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 29732 Accepted: 10863

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

POJ Monthly,Lou Tiancheng

题意：给出一个n*n的01矩阵，操作1是将其中一个子矩阵全部取反，操作2是查询矩阵内某一元素的值

这题我们并不方便直接将每次操作更新到叶子节点，也不需要打lazy标记，只需要在查询的时候将从根节点到我们目标节点一路上被更新的次数相加%2

#include <stdio.h>using namespace std;const int N = 1e3 + 10;int t, n, m;int T[N<<2][N<<2];void Build2(int l, int r, int deep, int k){    T[deep][k] = 0;    if(l == r) return;    int mid = (l+r)>>1;    Build2(l, mid, deep, k<<1);    Build2(mid+1, r, deep, k<<1|1);}void Build(int l, int r, int k){    Build2(1, n, k, 1);    if(l == r) return;    int mid = (l+r)>>1;    Build(l, mid, k<<1);    Build(mid+1, r, k<<1|1);}void Update2(int l, int r, int L, int R, int deep, int k){    if(l <= L && r >= R){        T[deep][k]++;        return;    }    int mid = (L+R)>>1;    if(r <= mid) Update2(l, r, L, mid, deep, k<<1);    else if(l > mid) Update2(l, r, mid+1, R, deep, k<<1|1);    else{        Update2(l, mid, L, mid, deep, k<<1);        Update2(mid+1, r, mid+1, R, deep, k<<1|1);    }}void Update(int l, int r, int L, int R, int ll, int rr, int k){    if(l <= L && r >= R){        Update2(ll, rr, 1, n, k, 1);        return;    }    int mid = (L+R)>>1;    if(r <= mid) Update(l, r, L, mid, ll, rr, k<<1);    else if(l > mid) Update(l, r, mid+1, R, ll, rr, k<<1|1);    else{        Update(l, mid, L, mid, ll, rr, k<<1);        Update(mid+1, r, mid+1, R, ll, rr, k<<1|1);    }}int ans;void Query2(int y, int L, int R, int deep, int k){    ans += T[deep][k];    if(L == R) return;    int mid = (L+R)>>1;    if(y <= mid) Query2(y, L, mid, deep, k<<1);    else Query2(y, mid+1, R, deep, k<<1|1);}void Query(int x, int y, int L, int R, int k){    Query2(y, 1, n, k, 1);    if(L == R) return;    int mid = (L+R)>>1;    if(x <= mid) Query(x, y, L, mid, k<<1);    else Query(x, y, mid+1, R, k<<1|1);}int main(){    scanf("%d", &t);    while(t--){        scanf("%d%d", &n, &m);        Build(1, n, 1);        char op[3];        int x1, y1, x2, y2;        while(m--){            scanf("%s", op);            if(op[0] == 'C'){                scanf("%d%d%d%d", &x1, &y1, &x2, &y2);                Update(x1, x2, 1, n, y1, y2, 1);            }            else{                scanf("%d%d", &x1, &y1);                ans = 0;                Query(x1, y1, 1, n, 1);                printf("%d\n", ans&1);            }        }        if(t) printf("\n");    }}