• 二维树状数组(单点更新+区间查询)


    http://poj.org/problem?id=1195

    题意:对矩阵进行操作。

     https://www.cnblogs.com/aininot260/p/9336527.html

    //#include<bits/stdc++.h>
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    #include <iostream>
    #include <string>
    #include <stdio.h>
    #include <queue>
    #include <stack>
    #include <map>
    #include <set>
    #include <string.h>
    #include <vector>
    typedef long long ll ;
    #define int ll
    #define mod 1000000007
    #define gcd __gcd
    #define rep(i , j , n) for(int i = j ; i <= n ; i++)
    #define red(i , n , j)  for(int i = n ; i >= j ; i--)
    #define ME(x , y) memset(x , y , sizeof(x))
    //ll lcm(ll a , ll b){return a*b/gcd(a,b);}
    //ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
    //int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
    //const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
    #define SC scanf
    #define INF  0x3f3f3f3f
    #define PI acos(-1)
    #define pii pair<int,int>
    #define fi first
    #define se second
    #define lson l,mid,root<<1
    #define rson mid+1,r,root<<1|1
    using namespace std;
    const int N = 1e6+100;
    const int maxn = 2e3+9;
    int c[maxn][maxn];
    int n ;
    
    int lowerbit(int x){
        return x&(-x);
    }
    
    
    void add(int x , int y , int val){
        while(x <= n){
            int j = y ;
            while(j <= n){
                c[x][j] += val;
                j += lowerbit(j);
            }
            x += lowerbit(x);
        }
    }
    int getsum(int x , int y){
        int ans = 0 ;
        while(x){
            int j = y ;
            while(j){
                ans += c[x][j] ;
                j -= lowerbit(j);
            }
            x -= lowerbit(x);
        }
        return ans ;
    }
    
    
    void solve(){
        int t ;
        while(~scanf("%lld" , &t) && t <= 2){
            if(t == 0){
                cin >> n ;
            }else if(t == 1){
                int x , y , val;
                scanf("%lld%lld%lld" , &x , &y , &val);
                x++ , y++;
                add(x , y , val);
            }else{
                int x1 , x2 , y1 , y2 ;
                scanf("%lld%lld%lld%lld" , &x1 , &y1 , &x2 , &y2);
                x1++ , y1++ , x2++ , y2++;
                cout << getsum(x2 , y2) - getsum(x1-1 , y2) - getsum(x2 , y1-1) + getsum(x1-1 , y1-1) << endl;
            }
        }
    }
    
    signed main()
    {
        //ios::sync_with_stdio(false);
        //cin.tie(0); cout.tie(0);
        //cnt = 0;
        //int t ;
        //scanf("%lld" , &t);
        //while(t--){
            solve();
        //}
    }
    
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  • 原文地址:https://www.cnblogs.com/nonames/p/12391240.html
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