Balanced Lineup 线段树 rmq 区间最值

Balanced Lineup

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.

Input

Line 1: Two space-separated integers, N and Q. 
Lines 2.. N+1: Line i+1 contains a single integer that is the height of cow i 
Lines N+2.. N+ Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive.

Output

Lines 1.. Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.

Sample Input

6 31734251 54 62 2

Sample Output

630

这题是让求区间内的最大值与最小值的差  用RMQ 和线段树都行的；

RMQ

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<math.h>
using namespace std;
int n;
int dp[50010][100];
int dq[50010][100];
void rmq()
{
    for(int j=1;(1<<j)<=n+1; j++)
    {
        for(int i=1;i+(1<<j)-1<=n;i++)
        {
            dp[i][j]=min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
            dq[i][j]=max(dq[i][j-1],dq[i+(1<<(j-1))][j-1]);
        }
    }
}
int rmq1(int l,int r)//最小值 
{
    int k=(int)(log(r-l+1+0.0)/log(2.0));  
    return min(dp[l][k],dp[r-(1<<k)+1][k]);
}
int rmq2(int l,int r)//最大值 
{
    int k=(int)(log(r-l+1+0.0)/log(2.0));  
    return max(dq[l][k],dq[r-(1<<k)+1][k]);
}
int main()
{
int q;
    while(~scanf("%d%d",&n,&q))
    {
    for(int j=1; j<=n; j++)
    {
            scanf("%d",&dp[j][0]);
            dq[j][0]=dp[j][0];
    }
    rmq();
    while(q--)
    {
           int a,b;
           scanf("%d%d",&a,&b);
           printf("%d\n",rmq2(a,b)-rmq1(a,b));
    }
    }
    return 0;
}

线段树  

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define INF 0x3f3f3f
#define L i<<1,l,mid
#define R i<<1|1,mid+1,r
using namespace std;
int n;
int t[50000<<2],t2[50000<<2];
void creat(int i,int l,int r)
{
    if(l==r)
    {
        scanf("%d",&t[i]);
        t2[i]=t[i];
        return ;
    }
    int mid=(l+r)>>1;
    creat(L);
    creat(R);
    t[i]=min(t[i<<1],t[i<<1|1]);
    t2[i]=max(t2[i<<1],t2[i<<1|1]);
}
int cha(int i,int l,int r,int x,int y)
{
    if(x<=l&&r<=y) return t[i];
    int mid=(l+r)>>1;
    int res=INF;
    if(x<=mid) res=min(cha(L,x,y),res);
    if(mid<y) res=min(cha(R,x,y),res);
    return res;
}
int cha1(int i,int l,int r,int x,int y)
{
    if(x<=l&&r<=y) return t2[i];
    int mid=(l+r)>>1;
    int res=-1;
    if(x<=mid) res=max(cha1(L,x,y),res);
    if(mid<y) res=max(cha1(R,x,y),res);
    return res;
}
int main()
{
    int q;
    scanf("%d",&n);
    scanf("%d",&q);
    creat(1,1,n);
    while(q--)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        printf("%d\n",cha1(1,1,n,a,b)-cha(1,1,n,a,b));
    }
    return 0;
}

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