poj 2155 二维树状数组 二维线段树

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

二维树状数组。两个for循环就好

#include <cstdio>#include <string>#include <cstring>#include <iostream>#include <vector>#define lowbit(x) ((x)&(-x))using namespace std;const int maxn=1010;int c[maxn][maxn];int n;void update(int x,int y){    int i,k;    for(i=x;i<=n;i+=lowbit(i))    {        for(k=y;k<=n;k+=lowbit(k))            c[i][k]++;    }}int Get(int x,int y){       int i,k,sum=0;       for(i=x;i>0;i-=lowbit(i))       {        for(k=y;k>0;k-=lowbit(k))            sum+=c[i][k];       }           return sum;}int main(){    int t;    int x1,x2,y1,y2;     scanf("%d",&t);        int m;        while(t--)        {            memset(c,0,sizeof(c));            scanf("%d%d",&n,&m);            while(m--)            {                char ch;                scanf(" %c",&ch);                if(ch=='C')                 {                    scanf("%d%d%d%d",&x1,&y1,&x2,&y2);                    x1++,x2++,y1++,y2++;  //因为更新的时候下面的坐标有的减一所以得从二开始                    update(x2,y2);                    update(x1-1,y1-1);                     update(x1-1,y2);                    update(x2,y1-1);                }                else                {                    scanf("%d%d",&x1,&y1);                    printf("%d\n",Get(x1,y1)%2);                }            }            printf("\n");        }}

#include <iostream>#include <cstdio>#include <cstring>using namespace std;#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++;//总的全部面积有没有被翻过，就是父区间是否被翻过，先统计父区间                      //再统计子区间。    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(){    int t;    while(~scanf("%d",&t))    while(t--)    {        scanf("%d%d",&N,&T);        mem(tree);        for(int i=0;i<T;i++)        {            scanf(" %c%d%d",&ch,&X1,&Y1);            if(ch=='C')             {                scanf("%d%d",&X2,&Y2);                editX(1,1,N);            }            else            {                num=0;                queryX(1,1,N);                if(num&1) printf("1\n");                else printf("0\n");            }        }        if(t) printf("\n");     }}