BZOJ 1452: [JSOI2009]Count 二维树状数组

1452: [JSOI2009]Count

Time Limit: 1 Sec  

Memory Limit: 256 MB

题目连接

http://www.lydsy.com/JudgeOnline/problem.php?id=1452

Description

Input

Output

Sample Input

Sample Output

1

2

HINT

 

题意

 

题解：

裸的二维树状数组

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <bitset>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 301
#define mod 1001
#define eps 1e-9
#define pi 3.1415926
int Num;
//const int inf=0x7fffffff;
const ll inf=999999999;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//*************************************************************************************

int S[101][maxn][maxn];
int col[maxn][maxn];
int n,m;
int lowbit(int x){return x&(-x);}
void updata(int c[maxn][maxn],int x,int y,int w)
{
    for(int i=x;i<=n;i+=lowbit(i))
        for(int j=y;j<=m;j+=lowbit(j))
            c[i][j]+=w;
}
int sum(int c[maxn][maxn],int x,int y)
{
    int ans=0;
    for(int i=x;i;i-=lowbit(i))
        for(int j=y;j;j-=lowbit(j))
            ans+=c[i][j];
    return ans;
}

int solve(int c[maxn][maxn],int x1,int y1,int x2,int y2)
{
    return sum(c,x2,y2)-sum(c,x2,y1-1)-sum(c,x1-1,y2)+sum(c,x1-1,y1-1);
}

int main()
{
    n=read(),m=read();
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            int x=read();
            col[i][j]=x;
            updata(S[x],i,j,1);
        }
    }
    int q=read();
    while(q--)
    {
        int op=read();
        if(op==1)
        {
            int x=read(),y=read(),c=read();
            updata(S[col[x][y]],x,y,-1);
            col[x][y]=c;
            updata(S[col[x][y]],x,y,1);
        }
        else
        {
            int x1=read(),x2=read(),y1=read(),y2=read(),c=read();
            printf("%d\n",solve(S[c],x1,y1,x2,y2));
        }
    }
}