poj 3264RMQ问题（线段树，ST算法）

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 34766		Accepted: 16368
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


#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long LL;
const int MAX=0xfffffff;
int n;
const int mx=50050;
int num[mx];
class node{
public:
    int l,r,maxv,minv;
}tree[mx*4];
void pushup(int v)
{
    tree[v].maxv=max(tree[v*2].maxv,tree[v*2+1].maxv);
    tree[v].minv=min(tree[v<<1].minv,tree[v<<1|1].minv);
}
void build(int v,int l,int r)
{
    tree[v].l=l,tree[v].r=r;
    if(l==r)
    {
        tree[v].maxv=num[l];
        tree[v].minv=num[l];
        return ;
    }
    int mid=(l+r)/2;
    build(v*2,l,mid),build(v*2+1,mid+1,r);
    pushup(v);
}
int query1(int v,int l,int r)
{
    if(l==tree[v].l&&tree[v].r==r)  return tree[v].maxv;
    else
    {
        int mid=(tree[v].l+tree[v].r)/2;
        if(r<=mid)  return query1(v*2,l,r);    //如果不添加等号，程序无法运行
        else if(l>mid)  return query1(v*2+1,l,r);    //如果加上=，就发生结果变化了
        else   return max(query1(v*2,l,mid),query1(v*2+1,mid+1,r));
    }
}
int query2(int v,int l,int r)
{
    if(l==tree[v].l&&tree[v].r==r)  return tree[v].minv;
    else
    {
        int mid=(tree[v].l+tree[v].r)/2;
        if(r<=mid)  return query2(v*2,l,r);    //如果不添加等号，程序无法运行
        else if(l>mid)  return query2(v*2+1,l,r);    //如果加上=，就发生结果变化了
        else   return min(query2(v*2,l,mid),query2(v*2+1,mid+1,r));
    }
}
int main( )
{
    //freopen("1.txt","r",stdin);
    int q;
    while(scanf("%d %d",&n,&q)!=EOF)
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&num[i]);
        build(1,1,n);
        while(q--)
        {
            int a,b;
            scanf("%d %d",&a,&b);
            int ma=query1(1,a,b);
            int mi=query2(1,a,b);
            printf("%d\n",ma-mi);
        }
    }
    return 0;
}


#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<cmath>
using namespace std;
typedef long long LL;
const int MAX=0xfffffff;

const int mx=50050;
int ma[mx][20], mi[mx][20];
int n;
void  RMQ_ST()
{
    int lg=floor(log10(double(n))/log10(double(2)));
    for(int j=1;j<=lg;j++)
        for(int i=1;i<=n+1-(1<<j);i++)
        {
            ma[i][j] = max(ma[i][j-1], ma[i+(1<<(j-1))][j-1]);
            mi[i][j] = min(mi[i][j-1], mi[i+(1<<(j-1))][j-1]);
        }
}
int main( )
{
    int q;
    while(scanf("%d %d",&n,&q)!=EOF)
    {
        for(int i=1;i<=n;i++)
        {
            int a;
            scanf("%d",&a);
            ma[i][0]=mi[i][0]=a;
        }
        RMQ_ST();
        while(q--)
        {
            int a,b;
            scanf("%d %d",&a,&b);
            if(a>b)  a=a^b^(b=a);
            int lg=floor(log10(double(b-a+1))/log10(double(2)));
            printf("%d\n",max(ma[a][lg],ma[b-(1<<lg)+1][lg])-min(mi[a][lg],mi[b-(1<<lg)+1][lg]));
        }
    }
}