【BZOJ1047】[HAOI2007]理想的正方形

題面

bzoj
洛谷

題解

二維\(st\)表，程式碼是以前的

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<algorithm>
#include<ctime>
#define RG register
#define file(x) freopen(x".in","r",stdin);
using namespace std;

#define ll long long
inline int gi(){
    int data=0,w=1;
    char ch=0;
    while(ch!='-'&&(ch>'9'||ch<'0')) ch=getchar();
    if(ch=='-') w=-1,ch=getchar();
    while(ch>='0'&&ch<='9') data=data*10+ch-'0',ch=getchar();
    return w*data;
}

#define N 1010
int a,b,n;
int val[N][N],dp1[N][N][11],dp2[N][N][11],lg[N];
int MX(int e,int f,int g,int h){
    return max(max(e,f),max(g,h));
}
int MI(int e,int f,int g,int h){
    return min(min(e,f),min(g,h));
}
void Pre(){
    for(RG int i=1;i<=a;i++)
      for(RG int j=1;j<=b;j++)
        dp1[i][j][0]=dp2[i][j][0]=val[i][j];
        
    for(RG int k=1;k<=10;k++)
      for(RG int i=1;i<=a-(1<<k-1);i++)
        for(RG int j=1;j<=b-(1<<k-1);j++){
        dp1[i][j][k]=MX (dp1[i][j][k-1],
                         dp1[i+(1<<k-1)][j][k-1],
                         dp1[i][j+(1<<k-1)][k-1],
                         dp1[i+(1<<k-1)][j+(1<<k-1)][k-1]);
                         
        dp2[i][j][k]=MI (dp2[i][j][k-1],
                         dp2[i+(1<<k-1)][j][k-1],
                         dp2[i][j+(1<<k-1)][k-1],
                         dp2[i+(1<<k-1)][j+(1<<k-1)][k-1]);
    }
}
int query(int i,int j,int dis){
    int t=log(n)/log(2);
    int sum;
    sum=(MX(dp1[i][j][t],dp1[n+i-(1<<t)][j][t],dp1[i][j+n-(1<<t)][t],dp1[i+n-(1<<t)][j+n-(1<<t)][t])-
         MI(dp2[i][j][t],dp2[i+n-(1<<t)][j][t],dp2[i][j+n-(1<<t)][t],dp2[i+n-(1<<t)][j+n-(1<<t)][t]));
    return sum;
}
int main(){
    a=gi();b=gi();n=gi();
    for(RG int i=1;i<=a;i++)
      for(RG int j=1;j<=b;j++)
        val[i][j]=gi();
    Pre();
    
    int MIN=0x7fffffff;;
    for(RG int i=1;i<=a-n+1;i++)
       for(RG int j=1;j<=b-n+1;j++)
           MIN=min(MIN,query(i,j,n));
    printf("%d\n",MIN);
    return 0;
}