题目：

http://poj.org/problem?id=2155

题意：

有一个初始元素均为0的n∗n矩阵，有下面两种操作：

  1. C x1 y1 x2 y2 以x1 y1为左上角，以x2 y2为右上角，把在这个范围内的矩阵元素取反，0变1，0变0
  2. Q x y查询矩阵中x y位置的元素

思路：

二维线段树题目，二维树状数组也可以。用一个线段树管理行，这个线段树内的点都是一个线段树，管理行内相应的列。看了好多二维线段树代码，想找一个通用式的模板，奈何大家写的基本都不怎么通用，自己尝试着写了一个，在第二维线段树里面进行lazy操作，比较慢。

//1100ms
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 1000 + 10, INF = 0x3f3f3f3f;

bool g[N<<2][N<<2];
int n, m;
void updatey(int ly, int ry, int L, int R, int ky, int kx)
{
    if(ly <= L && R <= ry)
    {
        g[kx][ky] ^= 1; return;
    }
    int mid = (L + R) >> 1;
    if(ly <= mid) updatey(ly, ry, L, mid, ky << 1, kx);
    if(ry > mid) updatey(ly, ry, mid + 1, R, ky << 1|1, kx);
}
void updatex(int lx, int rx, int ly, int ry, int L, int R, int kx)
{
    if(lx <= L && R <= rx)
    {
        updatey(ly, ry, 1, n, 1, kx); return;
    }
    int mid = (L + R) >> 1;
    if(lx <= mid) updatex(lx, rx, ly, ry, L, mid, kx << 1);
    if(rx > mid) updatex(lx, rx, ly, ry, mid + 1, R, kx << 1|1);
}
int queryy(int y, int L, int R, int ky, int kx)
{
    int ans = g[kx][ky];
    if(L == R) return ans;
    int mid = (L + R) >> 1;
    if(y <= mid) ans ^= queryy(y, L, mid, ky << 1, kx);
    else ans ^= queryy(y, mid + 1, R, ky << 1|1, kx);
    return ans;
}
int queryx(int x, int y, int L, int R, int kx)
{
    int ans = queryy(y, 1, n, 1, kx);
    if(L == R) return ans;
    int mid = (L + R) >> 1;
    if(x <= mid) ans ^= queryx(x, y, L, mid, kx << 1);
    else ans ^= queryx(x, y, mid + 1, R, kx << 1|1);
    return ans;
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        memset(g, 0, sizeof g);
        scanf("%d%d", &n, &m);
        int x1, y1, x2, y2;
        char op;
        for(int i = 1; i <= m; i++)
        {
            scanf(" %c", &op);
            if(op == 'C')
            {
                scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
                updatex(x1, x2, y1, y2, 1, n, 1);
            }
            else
            {
                scanf("%d%d", &x1, &y1);
                printf("%d\n", queryx(x1, y1, 1, n, 1));
            }
        }
        if(t) printf("\n");
    }
    return 0;
}

自己写的通用式模板（其实也不怎么通用）：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 1000 + 10, INF = 0x3f3f3f3f;

bool val[N<<2][N<<2], lazy[N<<2][N<<2];
int n, m;

void push_downy(int ky, int kx)
{
    if(lazy[kx][ky])
    {
        val[kx][ky<<1] ^= lazy[kx][ky], lazy[kx][ky<<1] ^= lazy[kx][ky];
        val[kx][ky<<1|1] ^= lazy[kx][ky], lazy[kx][ky<<1|1] ^= lazy[kx][ky];
        lazy[kx][ky] = 0;
    }
}
void updatey(int ly, int ry, int L, int R, int ky, int kx)
{
    if(ly <= L && R <= ry)
    {
        val[kx][ky] ^= 1;
        lazy[kx][ky] ^= 1;
        return;
    }
    push_downy(ky, kx);
    int mid = (L + R) >> 1;
    if(ly <= mid) updatey(ly, ry, L, mid, ky << 1, kx);
    if(ry > mid) updatey(ly, ry, mid + 1, R, ky << 1|1, kx);
}
void updatex(int lx, int rx, int ly, int ry, int L, int R, int kx)
{
    if(lx <= L && R <= rx)
    {
        updatey(ly, ry, 1, n, 1, kx); return;
    }
    int mid = (L + R) >> 1;
    if(lx <= mid) updatex(lx, rx, ly, ry, L, mid, kx << 1);
    if(rx > mid) updatex(lx, rx, ly, ry, mid + 1, R, kx << 1|1);
}
int queryy(int y, int L, int R, int ky, int kx)
{
    int ans = 0;
    if(L == R)
    {
        return ans ^= val[kx][ky];
    }
    push_downy(ky, kx);
    int mid = (L + R) >> 1;
    if(y <= mid) ans ^= queryy(y, L, mid, ky << 1, kx);
    else ans ^= queryy(y, mid + 1, R, ky << 1|1, kx);
    return ans;
}
int queryx(int x, int y, int L, int R, int kx)
{
    int ans = queryy(y, 1, n, 1, kx);
    if(L == R) return ans;
    int mid = (L + R) >> 1;
    if(x <= mid) ans ^= queryx(x, y, L, mid, kx << 1);
    else ans ^= queryx(x, y, mid + 1, R, kx << 1|1);
    return ans;
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        memset(val, 0, sizeof val);
        memset(lazy, 0, sizeof lazy);
        scanf("%d%d", &n, &m);
        char op;
        int x1, y1, x2, y2;
        for(int i = 1; i <= m; i++)
        {
            scanf(" %c", &op);
            if(op == 'C')
            {
                scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
                updatex(x1, x2, y1, y2, 1, n, 1);
            }
            else
            {
                scanf("%d%d", &x1, &y1);
                printf("%d\n", queryx(x1, y1, 1, n, 1));
            }
        }
        if(t) printf("\n");
    }
    return 0;
}