POJ 3264——Balanced Lineup（RMQ ，segment tree，树状数组）

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 34288		Accepted: 16123
Case Time Limit: 2000MS

Description

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 3
1
7
3
4
2
5
1 5
4 6
2 2

Sample Output

6
3
0

题目大意：

求区间 [ L , R ] 最大值与最小值的差

RMQ代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#define M 50000+10
using namespace std;
int a[M],dp1[M][16],dp2[M][16];
int n,m;
void RMQ_init()
{
    for(int i=1;i<=n;++i) dp1[i][0]=dp2[i][0]=a[i];
    for(int j=1;(1<<j)<=n;++j){
        for(int i=1;i+(1<<j)-1<=n;++i){
            dp1[i][j]=min(dp1[i][j-1],dp1[i+(1<<(j-1))][j-1]);
            dp2[i][j]=max(dp2[i][j-1],dp2[i+(1<<(j-1))][j-1]);
        }
    }
}
int RMQ(int l,int r)
{
    int k=(int)(log(r-l+1.0)/log(2.0));
    return max(dp2[l][k],dp2[r-(1<<k)+1][k])-min(dp1[l][k],dp1[r-(1<<k)+1][k]);
}
int main()
{
    scanf("%d %d",&n,&m);
    for(int i=1;i<=n;++i){
        scanf("%d",&a[i]);
    }
    RMQ_init();
    int l,r;
    while(m--){
        scanf("%d %d",&l,&r);
        printf("%d\n",RMQ(l,r));
    }
    return 0;
}

Segment Tree代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define M 50000+10
#define l(x) x<<1
#define r(x) x<<1|1
using namespace std;
int a[M];
int n,m;
int Max,Min;
struct node
{
    int l,r,minn,maxx,dif;
}st[M<<2];

void push_up(int u)
{
    st[u].minn=min( st[ l(u) ].minn, st[ r(u) ].minn );
    st[u].maxx=max( st[ l(u) ].maxx, st[ r(u) ].maxx );
}

void build(int u,int l,int r)
{
    st[u].l=l,st[u].r=r;st[u].minn=st[u].maxx=0;
    if(l==r){
        st[u].minn=st[u].maxx=a[st[u].l];
        return ;
    }
    int m=(l+r)>>1;
    build(l(u),l,m);
    build(r(u),m+1,r);
    push_up(u);
}

/*void insert(int u,int p,int v)
{
    if(st[u].l==st[u].r){
        st[u].minn=st[u].maxx=v;
        return ;
    }
    int m=(st[u].l+st[u].r)>>1;
    if(p<=m) insert(l(u),p,v);
    else insert(r(u),p,v);
    push_up(u);
}*/
void query(int u,int i,int j)
{
    if(i<=st[u].l&&st[u].r<=j){
        Max=max(Max,st[u].maxx);
        Min=min(Min,st[u].minn);
        return ;
    }
    int m=(st[u].l+st[u].r)>>1;
    if(i<=m) query(l(u),i,j);
    if(m<j) query(r(u),i,j);
}
int main()
{

    scanf("%d %d",&n,&m);

    int x;
    for(int i=1;i<=n;++i){
        scanf("%d",&a[i]);
    }
    build(1,1,n);
    while(m--){
        int l,r;
        scanf("%d %d",&l,&r);
        Max=-1<<30,Min=1<<30;
        query(1,l,r);
        printf("%d\n",Max-Min);
    }
    return 0;
}

树状数组代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define M 50000+10
using namespace std;
int n,m,a[M];
int minv[M],maxv[M];
int lowbit(int x){return x&-x;}
void update(int x,int v)
{
    while(x<=n){
        minv[x]=min(minv[x],v);
        maxv[x]=max(maxv[x],v);
        x+=lowbit(x);
    }
}
int query(int l,int r)
{
    int Max=-1<<30,Min=1<<30;
    while(r>=l){
        int low=lowbit(r);
        if(r-low>=l){
            Max=max(Max,maxv[r]);
            Min=min(Min,minv[r]);
            r-=low;
        }
        else{
            Max=max(Max,a[r]);
            Min=min(Min,a[r]);
            r--;
        }
    }
    return Max-Min;
}

int main()
{
    scanf("%d %d",&n,&m);
    memset(minv,0x3f,sizeof minv);
    memset(maxv,-1,sizeof maxv);
    for(int i=1;i<=n;++i){
        scanf("%d",&a[i]);
        update(i,a[i]);
    }
    while(m--){
        int l,r;
        scanf("%d %d",&l,&r);
        printf("%d\n",query(l,r));
    }

    return 0;
}

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