POJ - 2155 Matrix （二维线段树/二维树状数组）

题目链接
题意：
两种操作：
\(1. C\) \(x_1\) \(y_1\) \(x_2\) \(y_2\)，将右上点为\((x_1, y_1)\),左下点为\((x_2, y_2)\)内的值全都异或1。
\(2. Q\) \(x\) \(y\)，查询点(\(x\), \(y\))的值。

思路：
\(1.\)二维线段树区间更新+单点查询。
与一维线段树区间更新不同，二维区间更新时懒惰lazy标记不需要下放，查询时需将包含该点的区间的懒惰标记全都记录。

\(2.\)二维树状数组区间更新+单点查询。
二维树状数组区间更新利用二维数组差分，查询点值即为求二维前缀和。

code:
二维线段树（常数较大）：

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <deque>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
// #include <unordered_map>

#define fi first
#define se second
#define pb push_back
#define endl "\n"
#define debug(x) cout << #x << ":" << x << endl;
#define bug cout << "********" << endl;
#define all(x) x.begin(), x.end()
#define lowbit(x) x & -x
#define fin(x) freopen(x, "r", stdin)
#define fout(x) freopen(x, "w", stdout)
#define ull unsigned long long
#define ll long long 

const double eps = 1e-15;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const double pi = acos(-1.0);
const int mod =  1e9 + 7;
const int maxn = 1e3 + 10;

using namespace std;

#define lson rt << 1 
#define rson rt << 1 | 1

int t[maxn << 2][maxn << 2], n;

void rebuild(int rt, int l, int r, int num){
    t[num][rt] = 0;
    if(l == r)return ;
    int mid = l + r >> 1;
    rebuild(lson, l, mid, num), rebuild(rson, mid + 1, r, num);
}

void build(int rt, int l, int r){
    rebuild(1, 1, n, rt);
    if(l == r)return;
    int mid = l + r >> 1;
    build(lson, l, mid), build(rson, mid + 1, r);
}

void reupdate(int rt, int l, int r, int Y1, int Y2, int num){
    if(Y1 <= l && r <= Y2)return void(t[num][rt] ^= 1);
    int mid = l + r >> 1;
    if(Y1 <= mid)reupdate(lson, l, mid, Y1, Y2, num);
    if(mid < Y2)reupdate(rson, mid + 1, r, Y1, Y2, num);
}

void update(int rt, int l, int r, int X1, int Y1, int X2, int Y2){
    if(X1 <= l && r <= X2)return void(reupdate(1, 1, n, Y1, Y2, rt));
    int mid = l + r >> 1;
    if(X1 <= mid)update(lson, l, mid, X1, Y1, X2, Y2);
    if(mid < X2)update(rson, mid + 1, r, X1, Y1, X2 ,Y2);
}

int requery(int rt, int l, int r, int Y1, int num){
    int ret = 0;
    ret ^= t[num][rt];
    if(l == r)return ret;
    int mid = l + r >> 1;
    if(Y1 <= mid)return ret ^ requery(lson, l, mid, Y1, num);
    else return ret ^ requery(rson, mid + 1, r, Y1, num);
}

int query(int rt, int l, int r, int X1, int Y1){
    int ret = 0;
    ret ^= requery(1, 1, n, Y1, rt);
    if(l == r)return ret;
    int mid = l + r >> 1;
    if(X1 <= mid)return ret ^ query(lson, l, mid, X1, Y1);
    else return ret ^ query(rson, mid + 1, r, X1, Y1);
}

int main(){
    int t, q, a1, b1, a2, b2;
    scanf("%d", &t);
    while(t --){
        scanf("%d%d", &n, &q);
        build(1, 1, n);
        char s[2];
        while(q --){
            scanf("%s", s);
            if(s[0] == 'C'){
                scanf("%d%d%d%d", &a1, &b1, &a2, &b2);
                update(1, 1, n, a1, b1, a2, b2);
            }
            else{
                scanf("%d%d", &a1, &b1);
                int ans = query(1, 1, n, a1, b1);
                printf("%d\n", ans);
            }
        }
        puts("");
    }
    return 0;
}

二维树状数组：

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <deque>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
// #include <unordered_map>

#define fi first
#define se second
#define pb push_back
#define endl "\n"
#define debug(x) cout << #x << ":" << x << endl;
#define bug cout << "********" << endl;
#define all(x) x.begin(), x.end()
#define lowbit(x) x & -x
#define fin(x) freopen(x, "r", stdin)
#define fout(x) freopen(x, "w", stdout)
#define ull unsigned long long
#define ll long long 

const double eps = 1e-15;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const double pi = acos(-1.0);
const int mod =  1e9 + 7;
const int maxn = 1e3 + 10;

using namespace std;

#define lson rt << 1 
#define rson rt << 1 | 1

int t[maxn][maxn], n;

void update(int x, int y){
    while(x <= n){
        int dy = y;
        while(dy <= n)t[x][dy] += 1, dy += lowbit(dy);
        x += lowbit(x);
    }
}

int getsum(int x, int y){
    int ret = 0;
    while(x){
        int dy = y;
        while(dy)ret += t[x][dy], dy -= lowbit(dy);
        x -= lowbit(x);
    }
    return ret;
}

int main(){
    int t1, q, a1, b1, a2, b2;
    scanf("%d", &t1);
    while(t1 --){
        memset(t, 0, sizeof t);
        scanf("%d%d", &n, &q);
        char s[2];
        while(q --){
            scanf("%s", s);
            if(s[0] == 'C'){
                scanf("%d%d%d%d", &a1, &b1, &a2, &b2);
                update(a1, b1), update(a2 + 1, b1), update(a1, b2 + 1), update(a2 + 1, b2 + 1);
            }
            else{
                scanf("%d%d", &a1, &b1);
                int ans = getsum(a1, b1) % 2;
                printf("%d\n", ans);
            }
        }
        puts("");
    }
    return 0;
}