[poj 2155]Matrix [2D BIT][二维线段树]

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

#include<cstdio>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<cctype>using namespace std;int T,n,m;char op[2];inline void read(int& x){char c = getchar();x = 0;while(!isdigit(c))c=getchar();while(isdigit(c)){x=x*10+c-'0';c=getchar();}}#define maxn 1005#define lowbit(x) (x&(-x))int c[maxn][maxn];inline void init(){memset(c,0,sizeof(c));}//其实直接加也可以不用弄成+1,-1(%2等价） inline void add(int x,int y,int d){for(int i=x;i<=n;i+=lowbit(i))for(int j=y;j<=n;j+=lowbit(j))c[i][j]+=d;}inline int sum(int x,int y){int ret = 0;for(int i=x;i>0;i-=lowbit(i))for(int j=y;j>0;j-=lowbit(j))ret+=c[i][j];return ret;}inline void update(int x1,int y1,int x2,int y2){add(x1-1,y1-1,1);add(x1-1,y2,-1);add(x2,y1-1,-1);add(x2,y2,1);}int main(void){read(T);int x1,y1,x2,y2;while(T--){read(n),read(m);init();for(int i=1;i<=m;i++){scanf("%s",op);if(*op=='C'){read(x1),read(y1),read(x2),read(y2);update(x1+1,y1+1,x2+1,y2+1);}else{read(x1),read(y1);printf("%d\n",sum(x1,y1)%2);}}puts("");}return 0;}

下面补个二维线段树

#include <stdio.h>#include <string.h>#define xlson kx<<1, xl, mid#define xrson kx<<1|1, mid+1, xr#define ylson ky<<1, yl, mid#define yrson ky<<1|1, mid+1, yr#define MAXN 1005#define mem(a) memset(a, 0, sizeof(a))bool tree[MAXN<<2][MAXN<<2];int  X, N, T;int num, X1, X2, Y1, Y2;char ch;void editY(int kx,int ky,int yl,int yr){    if(Y1<=yl && yr<=Y2)    {        tree[kx][ky] = !tree[kx][ky];        return ;    }    int mid = (yl+yr)>>1;    if(Y1 <= mid) editY(kx,ylson);    if(Y2 >  mid) editY(kx,yrson);}void editX(int kx,int xl,int xr){    if(X1<=xl && xr<=X2)    {        editY(kx,1,1,N);        return ;    }    int mid = (xl+xr)>>1;    if(X1 <= mid) editX(xlson);    if(X2 >  mid) editX(xrson);}void queryY(int kx,int ky,int yl,int yr){    if(tree[kx][ky]) num ++;    if(yl==yr) return ;    int mid = (yl+yr)>>1;    if(Y1 <= mid) queryY(kx,ylson);    else queryY(kx,yrson);}void queryX(int kx,int xl,int xr){    queryY(kx,1,1,N);    if(xl==xr) return ;    int mid = (xl+xr)>>1;    if(X1 <= mid)queryX(xlson);    else  queryX(xrson);}int main(){    while(~scanf("%d", &X))while(X--)    {        mem(tree);        scanf("%d %d%*c", &N,&T);        for(int i=0;i<T;i++)        {            scanf("%c %d %d%*c",&ch,&X1,&Y1);            if(ch == 'C')            {                scanf("%d %d%*c", &X2, &Y2);                editX(1,1,N);            }            else            {                num = 0;                queryX(1,1,N);                if(num & 1)printf("1\n");                else printf("0\n");            }        }        if(X) printf("\n");    }    return 0;}