詢問區間內最大值-最小值

POJ 3264 Balanced Lineup（簡單RMQ）

http://poj.org/problem?id=3264

題意：

 

        For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.

        Farmer John has made a list of Q (1 ≤ Q ≤ 200,000) potential groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.

分析：

        RMQ問題的簡單應用。

        即給你一個數組，要你輸出每次詢問區間內的最大值-最小值的差。

AC代碼：1641ms

<span style="font-size:18px;">#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=50000+1000;
int dmax[MAXN][20];
int dmin[MAXN][20];
int d[MAXN];
void initMax(int n,int d[])
{
    for(int i=1;i<=n;i++)dmax[i][0]=d[i];
    for(int j=1;(1<<j)<=n;j++)
        for(int i=1;i+(1<<j)-1<=n;i++)
            dmax[i][j]=max(dmax[i][j-1],dmax[i+(1<<(j-1))][j-1]);
}
int getMax(int L,int R)
{
    int k=0;
    while((1<<(k+1))<=R-L+1)k++;
    return max( dmax[L][k],dmax[R-(1<<k)+1][k] );
}
void initMin(int n,int d[])
{
    for(int i=1;i<=n;i++)dmin[i][0]=d[i];
    for(int j=1;(1<<j)<=n;j++)
        for(int i=1;i+(1<<j)-1<=n;i++)
            dmin[i][j]=min( dmin[i][j-1],dmin[i+(1<<(j-1))][j-1] );
}
int getMin(int L,int R)
{
    int k=0;
    while((1<<(k+1))<=R-L+1)k++;
    return min( dmin[L][k],dmin[R-(1<<k)+1][k] );
}
 
int main()
{
    int n,q;
    while(scanf("%d%d",&n,&q)==2)
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&d[i]);
        initMax(n,d);
        initMin(n,d);
        while(q--)
        {
            int L,R;
            scanf("%d%d",&L,&R);
            printf("%d\n",getMax(L,R)-getMin(L,R));
        }
    }
    return 0;
}
</span>