题目链接：<poj.org/problem?id=2155>
大意：一个初始值全为0的01矩阵，有两种操作，一种是将某个矩形区域的值都取反，另一种是查询某一单点的值。
受到另一题启发：https://seekstar.github.io/2019/07/08/codeforces-1184-c2-曼哈顿距离转切比雪夫距离，矩形覆盖最大点数转最大前缀/
我们定义另一个矩阵C，使得原矩阵的某一个点的值为C的前缀异或，那么如果我们要将原矩阵的左下角为(x1, y1), 右上角为(x2, y2)的矩形区域的值取反，只需要将C矩阵的点(x1, y1), (x1, y2 + 1), (x2 + 1, y1), (x2 + 1, y2 + 1)的值取反即可。
代码：
#include <cstdio>
#include <cctype>
using namespace std;
#define DEBUG 0
#define ONLINE_JUDGE
int lowbit(int x) {
return x & -x;
}
template <typename T>
struct BIT_2DIM {
const static int maxn1 = 1010, maxn2 = 1010;
T s[maxn1][maxn2];
int n1, n2;
void Init(int _n1, int _n2) {
n1 = _n1;
n2 = _n2;
for (int i = 0; i <= n1; ++i)
for (int j = 0; j <= n2; ++j)
s[i][j] = 0;
}
void Add(int x1, int x2, T v) {
for (int t = x2; x1 <= n1; x1 += lowbit(x1))
for (x2 = t; x2 <= n2; x2 += lowbit(x2))
s[x1][x2] ^= v;
}
T Sum(int x1, int x2) {
T ans = 0;
for (int t = x2; x1; x1 ^= lowbit(x1))
for (x2 = t; x2; x2 ^= lowbit(x2))
ans ^= s[x1][x2];
return ans;
}
};
int main() {
#ifndef ONLINE_JUDGE
freopen("in.in", "r", stdin);
freopen("out.out", "w", stdout);
#endif
int T;
static BIT_2DIM<bool> bit;
bool first = true;
scanf("%d", &T);
while (T--) {
if (first) first = false;
else putchar('\n');
int n, m;
scanf("%d%d", &n, &m);
bit.Init(n + 5, n + 5);
while (m--) {
getchar();
char op = getchar();
if ('C' == op) {
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
bit.Add(x1, y1, 1);
bit.Add(x1, y2 + 1, 1);
bit.Add(x2 + 1, y1, 1);
bit.Add(x2 + 1, y2 + 1, 1);
} else {
int x, y;
scanf("%d%d", &x, &y);
printf("%d\n", bit.Sum(x, y));
}
}
}
return 0;
}
另外有一个二维线段树写法，这里用永久标记实现
#include <cstdio>
#include <cstring>
#include <cctype>
using namespace std;
#define MAXN1 1011
#define MAXN2 1011
inline int ls(int rt) {
return rt << 1;
}
inline int rs(int rt) {
return rt << 1 | 1;
}
struct SGT {
const static int max_node = MAXN2 << 2;
bool s[max_node];
void Init(int n) {
memset(s, 0, (n<<2) * sizeof(s[0]));
}
void Reverse(int rt, int l, int r, int L, int R) {
if (L <= l && r <= R) {
s[rt] = !s[rt];
} else {
int mid = (l + r) >> 1;
if (L <= mid) Reverse(ls(rt), l, mid, L, R);
if (mid < R) Reverse(rs(rt), mid+1, r, L, R);
}
}
bool Query(int rt, int l, int r, int x) {
if (l == r) {
return s[rt];
} else {
int mid = (l + r) >> 1;
if (x <= mid) return s[rt] ^ Query(ls(rt), l, mid, x);
else return s[rt] ^ Query(rs(rt), mid+1, r, x);
}
}
};
struct SGT2 {
const static int max_node = MAXN1 << 2;
SGT s[max_node];
int n2;
void Init(int n1, int _n2) {
n2 = _n2;
for (int i = 0; i <= (n1 << 2); ++i)
s[i].Init(n2);
}
void Reverse(int rt, int l, int r, int L1, int R1, int L2, int R2) {
if (L1 <= l && r <= R1) {
s[rt].Reverse(1, 1, n2, L2, R2);
} else {
int mid = (l + r) >> 1;
if (L1 <= mid) Reverse(ls(rt), l, mid, L1, R1, L2, R2);
if (mid < R1) Reverse(rs(rt), mid+1, r, L1, R1, L2, R2);
}
}
bool Query(int rt, int l, int r, int x1, int x2) {
bool ans = s[rt].Query(1, 1, n2, x2);
if (l != r) {
int mid = (l + r) >> 1;
if (x1 <= mid) ans ^= Query(ls(rt), l, mid, x1, x2);
else ans ^= Query(rs(rt), mid+1, r, x1, x2);
}
return ans;
}
};
char GetChar() {
char ch;
while (isspace(ch = getchar()));
return ch;
}
int main() {
int T;
static SGT2 sgt2;
scanf("%d", &T);
bool first = true;
while (T--) {
if (first) first = false;
else putchar('\n');
int n, m;
scanf("%d%d", &n, &m);
sgt2.Init(n, n);
while (m--) {
char op = GetChar();
if (op == 'Q') {
int x, y;
scanf("%d%d", &x, &y);
putchar(sgt2.Query(1, 1, n, x, y) + '0');
putchar('\n');
} else {
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
sgt2.Reverse(1, 1, n, x1, x2, y1, y2);
}
}
}
return 0;
}
