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RMQ [POJ 2019] Cornfields

2024-07-28 20:24:46   223人阅读

Cornfields
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 5516 Accepted: 2714

Description

FJ has decided to grow his own corn hybrid in order to help the cows make the best possible milk. To that end, he‘s looking to build the cornfield on the flattest piece of land he can find. 

FJ has, at great expense, surveyed his square farm of N x N hectares (1 <= N <= 250). Each hectare has an integer elevation (0 <= elevation <= 250) associated with it. 

FJ will present your program with the elevations and a set of K (1 <= K <= 100,000) queries of the form "in this B x B submatrix, what is the maximum and minimum elevation?". The integer B (1 <= B <= N) is the size of one edge of the square cornfield and is a constant for every inquiry. Help FJ find the best place to put his cornfield. 

Input

* Line 1: Three space-separated integers: N, B, and K. 

* Lines 2..N+1: Each line contains N space-separated integers. Line 2 represents row 1; line 3 represents row 2, etc. The first integer on each line represents column 1; the second integer represents column 2; etc. 

* Lines N+2..N+K+1: Each line contains two space-separated integers representing a query. The first integer is the top row of the query; the second integer is the left column of the query. The integers are in the range 1..N-B+1. 

Output

* Lines 1..K: A single integer per line representing the difference between the max and the min in each query. 

Sample Input

5 3 15 1 2 6 31 3 5 2 77 2 4 6 19 9 8 6 50 6 9 3 91 2

Sample Output

5

Source

USACO 2003 March Green
 
矩形解法：
#include <iostream>#include <cstdio>#include <cmath>using namespace std;#define N 255int n,b,k;int val[N][N];int mx[N][N][8][8];int mi[N][N][8][8];void ST(int n,int m){    int i,j,r,c;    for(i=1;i<=n;i++)    {        for(j=1;j<=m;j++)        {            mx[i][j][0][0]=mi[i][j][0][0]=val[i][j];        }    }    int kn=(int)(log(double(n))/log(2.0));    int km=(int)(log(double(m))/log(2.0));    for(i=0;i<=kn;i++)    {        for(j=0;j<=km;j++)        {            if(i==0 && j==0)  continue;            for(r=1;r+(1<<i)-1<=n;r++)            {                for(c=1;c+(1<<j)-1<=m;c++)                {                    if(i==0)                    {                        mx[r][c][i][j]=max(mx[r][c][i][j-1],mx[r][c+(1<<(j-1))][i][j-1]);                        mi[r][c][i][j]=min(mi[r][c][i][j-1],mi[r][c+(1<<(j-1))][i][j-1]);                    }                    else                    {                        mx[r][c][i][j]=max(mx[r][c][i-1][j],mx[r+(1<<(i-1))][c][i-1][j]);                        mi[r][c][i][j]=min(mi[r][c][i-1][j],mi[r+(1<<(i-1))][c][i-1][j]);                    }                }            }        }    }}int RMQ(int r1,int c1,int r2,int c2){    int kr=(int)(log(double(r2-r1+1))/log(2.0));    int kc=(int)(log(double(c2-c1+1))/log(2.0));        int t1=mx[r1][c1][kr][kc];    int t2=mx[r2-(1<<kr)+1][c1][kr][kc];    int t3=mx[r1][c2-(1<<kc)+1][kr][kc];    int t4=mx[r2-(1<<kr)+1][c2-(1<<kc)+1][kr][kc];        int m1=mi[r1][c1][kr][kc];    int m2=mi[r2-(1<<kr)+1][c1][kr][kc];    int m3=mi[r1][c2-(1<<kc)+1][kr][kc];    int m4=mi[r2-(1<<kr)+1][c2-(1<<kc)+1][kr][kc];        return max(max(t1,t2),max(t3,t4))-min(min(m1,m2),min(m3,m4));}int main(){    int i,j;    scanf("%d%d%d",&n,&b,&k);    for(i=1;i<=n;i++)    {        for(j=1;j<=n;j++)        {            scanf("%d",&val[i][j]);        }    }    ST(n,n);    while(k--)    {        int r1,c1,r2,c2;        scanf("%d%d",&r1,&c1);        r2=r1+b-1;        c2=c1+b-1;        printf("%d\n",RMQ(r1,c1,r2,c2));    }    return 0;}

正方形解法

#include <stdio.h>#include <iostream>#include <math.h>using namespace std;#define inf 0x7fffffff#define N 255#define max(a,b) a>b?a:b#define min(a,b) a<b?a:bint n,b,k;int val[N][N];int mx[N][N][8];int mi[N][N][8];int getMax(int x,int y,int p){    int res=-inf;    res=max(res,mx[x][y][p]);    if(x+(1<<p)<=n) res=max(res,mx[x+(1<<p)][y][p]);    if(y+(1<<p)<=n) res=max(res,mx[x][y+(1<<p)][p]);    if(x+(1<<p)<=n && y+(1<<p)<=n) res=max(res,mx[x+(1<<p)][y+(1<<p)][p]);    return res;}int getMin(int x,int y,int p){    int res=inf;    res=min(res,mi[x][y][p]);    if(x+(1<<p)<=n) res=min(res,mi[x+(1<<p)][y][p]);    if(y+(1<<p)<=n) res=min(res,mi[x][y+(1<<p)][p]);    if(x+(1<<p)<=n && y+(1<<p)<=n) res=min(res,mi[x+(1<<p)][y+(1<<p)][p]);    return res;}void ST(){    int i,j,k;    for(i=1;i<=n;i++)    {        for(j=1;j<=n;j++)        {            mx[i][j][0]=mi[i][j][0]=val[i][j];        }    }    int kn=(int)(log(n*1.0)/log(2.0));        for(k=1;k<=kn;k++)     {        for(i=1;i<=n;i++)        {            for(j=1;j<=n;j++)            {                mx[i][j][k]=getMax(i,j,k-1);                mi[i][j][k]=getMin(i,j,k-1);            }        }    }}int RMQMAX(int x,int y,int b){    int p=(int)(log(b*1.0)/log(2.0));    int res=-inf;    res=max(res,mx[x][y][p]);    if(x+b-(1<<p)<=n) res=max(res,mx[x+b-(1<<p)][y][p]);    if(y+b-(1<<p)<=n) res=max(res,mx[x][y+b-(1<<p)][p]);    if(x+b-(1<<p)<=n && y+b-(1<<p)<=n) res=max(res,mx[x+b-(1<<p)][y+b-(1<<p)][p]);    return res;}int RMQMIN(int x,int y,int b){    int p=(int)(log(b*1.0)/log(2.0));    int res=inf;    res=min(res,mi[x][y][p]);    if(x+b-(1<<p)<=n) res=min(res,mi[x+b-(1<<p)][y][p]);    if(y+b-(1<<p)<=n) res=min(res,mi[x][y+b-(1<<p)][p]);    if(x+b-(1<<p)<=n && y+b-(1<<p)<=n) res=min(res,mi[x+b-(1<<p)][y+b-(1<<p)][p]);    return res;}int main(){    int i,j;    scanf("%d%d%d",&n,&b,&k);    for(i=1;i<=n;i++)    {        for(j=1;j<=n;j++)        {            scanf("%d",&val[i][j]);        }    }    ST();    while(k--)    {        int x,y;        scanf("%d%d",&x,&y);        printf("%d\n",RMQMAX(x,y,b)-RMQMIN(x,y,b));    }    return 0;}

 

RMQ [POJ 2019] Cornfields


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