二维树状数组模板（区间更新，单点查询）（以POJ 2155为例）

叁歲伎倆 2022-05-15 14:55 192阅读 0赞

## 题目：[点击打开链接][Link 1]   题意：n\*n坐标图起初都为0，C：翻转左下和右上两个坐标围成的矩阵中所有点，Q：查询此点的0 1状态。   分析：利用差分的思想，推广到二维，一维单点查询就是前缀和，即query(x)。区间修改先让s-n都加num，再让t+1-n减去num，即update(s, num)，update(t+1, -num)。二维的单点查询变成二维就好了query(x, y)。区间修改update(x1, y1, num), update(x2+1, y1, -num), update(x1, y2+1, -num), update(x2+1, y2+1, num)。统计翻转次数是偶次还是奇次就行了。参考博客[https://www.cnblogs.com/lesroad/p/8479839.html][https_www.cnblogs.com_lesroad_p_8479839.html]、[https://www.cnblogs.com/RabbitHu/p/BIT.html][https_www.cnblogs.com_RabbitHu_p_BIT.html]。 ##

## 代码： ##

#pragma comment(linker, "/STACK:102400000,102400000")
    ///#include<unordered_map>
    ///#include<unordered_set>
    #include<algorithm>
    #include<iostream>
    #include<fstream>
    #include<complex>
    #include<cstdlib>
    #include<cstring>
    #include<cassert>
    #include<iomanip>
    #include<string>
    #include<cstdio>
    #include<bitset>
    #include<vector>
    #include<cctype>
    #include<cmath>
    #include<ctime>
    #include<stack>
    #include<queue>
    #include<deque>
    #include<list>
    #include<set>
    #include<map>
    using namespace std;
    #define pt(a) cout<<a<<endl
    #define debug test
    #define mst(ss,b) memset((ss),(b),sizeof(ss))
    #define rep(i,a,n) for (int i=a;i<=n;i++)
    #define per(i,a,n) for (int i=n-1;i>=a;i--)
    #define pii pair<int,int>
    #define fi first
    #define se second
    #define ll long long
    #define ull unsigned long long
    #define pb push_back
    #define mp make_pair
    #define inf 0x3f3f3f3f
    #define eps 1e-10
    #define PI acos(-1.0)
    #define lb(x) (x&(-x))
    const ll mod = 1e9+7;
    const int N = 1e3+10;
    
    ll qp(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
    int to[4][2]={
       {-1,0},{1,0},{0,-1},{0,1}};
    
    int t,n,q,tr[N][N];
    
    void upd(int x,int y,int v) {
        for(int i=x;i<N;i+=lb(i))
            for(int j=y;j<N;j+=lb(j))
                tr[i][j]+=v;
    }
    
    int qy(int x,int y) {
        int res=0;
        for(int i=x;i>0;i-=lb(i))
            for(int j=y;j>0;j-=lb(j))
                res+=tr[i][j];
        return res;
    }
    
    int main() {
        ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
        while(~scanf("%d",&t)) {
            while(t--) {
                mst(tr,0);
                scanf("%d%d",&n,&q);
                while(q--) {
                    char s[10];
                    int x1,y1,x2,y2;
                    scanf("%s",s);
                    if(s[0]=='C') {
                        scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                        upd(x1,y1,1);
                        upd(x1,y2+1,-1);
                        upd(x2+1,y1,-1);
                        upd(x2+1,y2+1,1);
                    }else {
                        scanf("%d%d",&x1,&y1);
                        printf("%d\n",qy(x1,y1)&1);
                    }
                }
                printf("\n");
            }
        }
        return 0;
    }

[Link 1]: http://poj.org/problem?id=2155
[https_www.cnblogs.com_lesroad_p_8479839.html]: https://www.cnblogs.com/lesroad/p/8479839.html
[https_www.cnblogs.com_RabbitHu_p_BIT.html]: https://www.cnblogs.com/RabbitHu/p/BIT.html