POJ 2155 Matrix【二维线段树】

题目大意:给你一个全是０的Ｎ＊Ｎ矩阵，每次有两种操作:１将矩阵中一个子矩阵置反，２．查询某个点是０还是１

思路:裸的二维线段树

 

#include<iostream>
#include<cstdio>
#include <math.h>
#include<algorithm>
#include<string.h>
#include<queue>
#define MOD 1000003
#define maxn 4009
#define LL long long
using namespace std;
int n,num;
bool tree[maxn][maxn];
void add_y(int xx,int node,int l,int r,int ql,int qr)
{
        if(ql<=l && r<=qr){tree[xx][node]^=1;return ;}
        int mid=(l+r)>>1;
        if(ql<mid)add_y(xx,node*2,l,mid,ql,qr);
        if(qr>mid)add_y(xx,node*2+1,mid,r,ql,qr);
}
void add_x(int node,int l,int r,int ql,int qr,int y1,int y2)
{
        if(ql<=l && r<=qr){add_y(node,1,1,n+1,y1,y2);return ;}
        int mid=(l+r)>>1;
        if(ql<mid)add_x(node*2,l,mid,ql,qr,y1,y2);
        if(qr>mid)add_x(node*2+1,mid,r,ql,qr,y1,y2);
}
void query_y(int xx,int node,int l,int r,int y1)
{
        if(tree[xx][node]==1)num++;
        if(l+1==r)return ;
        int mid=(l+r)>>1;
        if(y1<mid)query_y(xx,node*2,l,mid,y1);
        else query_y(xx,node*2+1,mid,r,y1);
}
void query_x(int node,int l,int r,int x1,int y1)
{
        query_y(node,1,1,n+1,y1);
        if(l+1==r)return ;
        int mid=(l+r)>>1;
        if(x1<mid)query_x(node*2,l,mid,x1,y1);
        else query_x(node*2+1,mid,r,x1,y1);
}
int main()
{
        int t,m,x1,x2,y1,y2;
        char ch[10];
        scanf("%d",&t);
        while(t--)
        {
                memset(tree,0,sizeof(tree));
                scanf("%d%d",&n,&m);
                for(int i=1;i<=m;i++)
                {
                        scanf("%s",ch+1);
                        if(ch[1]=='C')
                        {
                                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                                add_x(1,1,n+1,x1,x2+1,y1,y2+1);
                        }
                        else
                        {
                                num=0;
                                scanf("%d%d",&x1,&y1);
                                query_x(1,1,n+1,x1,y1);
                                printf("%d\n",num&1);
                        }
                }
                printf("\n");
        }
        return 0;
}