POJ2019 Cornfields

题解:

二维RMQ中的ST算法的模板题

代码:

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<map>
#include<set>
#include<vector>
using namespace std;
using namespace std;
#define pb push_back
#define mp make_pair
#define se second
#define fs first
#define ll long long
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define ls o<<1
#define rs o<<1|1
#define SZ(x) ((int)(x).size())
#define FOR(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
typedef pair<int,int> P;
const double eps=1e-9;
const int maxn=50100;
const int N=1e9;
const int mod=1e9+7;

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 m[255][255];
int mi[255][255][8][8],mx[255][255][8][8];
int n,b,k,x,y;

void rmq_init()
{
    for(int i=1;i<=n;i++)
    for(int j=1;j<=n;j++) mi[i][j][0][0]=mx[i][j][0][0]=m[i][j];
    int Mx=(int)(log(n*1.0)/log(2.0));
    int My=(int)(log(n*1.0)/log(2.0));
    for(int i=0;i<=Mx;i++)
    for(int j=0;j<=My;j++)
    {
        if(i==0&&j==0) continue;
        for(int Row=1;Row+(1<<i)-1<=n;Row++)
        for(int Col=1;Col+(1<<j)-1<=n;Col++)
        {
            if(i)//因为已经算了只有一列时，每一一行的极值，那么有多列时，就不需要算行了
            {
                mx[Row][Col][i][j]=max(mx[Row][Col][i-1][j],mx[Row+(1<<(i-1))][Col][i-1][j]);
                mi[Row][Col][i][j]=min(mi[Row][Col][i-1][j],mi[Row+(1<<(i-1))][Col][i-1][j]);
            }
            else//当矩形只有一列时，那么算一行的极值
            {
                mx[Row][Col][i][j]=max(mx[Row][Col][i][j-1],mx[Row][Col+(1<<(j-1))][i][j-1]);
                mi[Row][Col][i][j]=min(mi[Row][Col][i][j-1],mi[Row][Col+(1<<(j-1))][i][j-1]);
            }
        }
    }
}

int rmq_max(int x1,int y1,int x2,int y2)
{
    int kx=(int)(log(x2-x1+1.0)/log(2.0));
    int ky=(int)(log(y2-y1+1.0)/log(2.0));
    x2=x2-(1<<kx)+1;
    y2=y2-(1<<ky)+1;
    return max(max(mx[x1][y1][kx][ky],mx[x1][y2][kx][ky]),max(mx[x2][y1][kx][ky],mx[x2][y2][kx][ky]));//算4部分，左右上下
}

int rmq_min(int x1,int y1,int x2,int y2)
{
    int kx=(int)(log(x2-x1+1.0)/log(2.0));
    int ky=(int)(log(y2-y1+1.0)/log(2.0));
    x2=x2-(1<<kx)+1;
    y2=y2-(1<<ky)+1;
    return min(min(mi[x1][y1][kx][ky],mi[x1][y2][kx][ky]),min(mi[x2][y1][kx][ky],mi[x2][y2][kx][ky]));
}

int main(){
    scanf("%d%d%d",&n,&b,&k);
    for(int i=1;i<=n;i++)
    for(int j=1;j<=n;j++) scanf("%d",&m[i][j]);
    rmq_init();
    while(k--){
        scanf("%d%d",&x,&y);
        printf("%d\n",rmq_max(x,y,x+b-1,y+b-1)-rmq_min(x,y,x+b-1,y+b-1));
    }
    return 0;
}