poj1195(二维树状数组)

题目链接:https://vjudge.net/problem/POJ-1195

题意:有s*s的矩阵,初始化为全0,有两种操作,单点修改(x,y)的值,区间查询(x,y)的值(l<=x<=r,b<=y<=t)。

思路:二维树状数组裸应用查询区间(l,b)~(r,t)的值可转换为tr[r][t]-tr[l-1][t]-tr[r][b-1]+tr[l-1][b-1]。要注意的是输入的x,y是从0开始的,所以要加1。

AC代码:  

#include<cstdio>
#include<cctype>
using namespace std;

inline int read(){
    int x=0,f=0;char c=0;
    while(!isdigit(c)){f|=c==‘-‘;c=getchar();}
    while(isdigit(c)){x=(x<<3)+(x<<1)+(c^48);c=getchar();}
    return f?-x:x;
}

int op,s,tr[1030][1030];

int lowbit(int x){
    return x&(-x);
}

void update(int x,int y,int k){
    while(x<=s){
        int tmp=y;
        while(tmp<=s){
            tr[x][tmp]+=k;
            tmp+=lowbit(tmp);
        }
        x+=lowbit(x);
    }
}

int query(int x,int y){
    int ans=0;
    while(x>0){
        int tmp=y;
        while(tmp>0){
            ans+=tr[x][tmp];
            tmp-=lowbit(tmp);
        }
        x-=lowbit(x);
    }
    return ans;
}

int main(){
    op=read(),s=read();
    while(op=read(),op!=3){
        if(op==1){
            int x=read()+1,y=read()+1,k=read();
            update(x,y,k);
        }
        else{
            int l=read()+1,b=read()+1,r=read()+1,t=read()+1;
            printf("%d\n",query(r,t)-query(l-1,t)-query(r,b-1)+query(l-1,b-1));
        }
    }
    return 0;
}

原文地址:https://www.cnblogs.com/FrankChen831X/p/10799465.html

时间: 2024-10-10 09:44:44

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