POJ 2155 Matrix 二维线段树

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;}