poj3264Balanced Lineup(线段树RMQ）

RMQ (Range Minimum/Maximum Query)问题是指：对于长度为n的数列A，回答若干询问RMQ(A,i,j)(i,j<=n)，返回数列A中下标在i,j里的最小(大）值，也就是说，RMQ问题是指求区间最值的问题。

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

1A 程序跑的好慢 3000+

输完更新 父节点的最小最大值 找的时候找两次 一次最大 一次最小 相减

#include <iostream>
#include<cstdio>
#include<string.h>
using namespace std;
#define N 50001
#define MAX 1000001
int s[N*4],mi[N*4],ma[N*4];
void upset(int l,int r,int w)
{
    if(l==r)
    return ;
    int m = (l+r)/2;
    upset(l,m,2*w);
    upset(m+1,r,2*w+1);
    mi[w] = min(mi[w*2],mi[2*w+1]);
    ma[w] = max(ma[w*2],ma[2*w+1]);
}
void build(int l,int r,int w)
{
    if(l==r)
    {
        scanf("%d", &s[w]);
        mi[w] = s[w];
        ma[w] = s[w];
        return ;
    }
    int m = (l+r)/2;
    build(l,m,2*w);
    build(m+1,r,2*w+1);
}
int search(int a,int b,int l,int r,int w)
{
    if(a==l&&b==r)
    return mi[w];
    int m = (l+r)/2,re = MAX;
    if(b<=m)
    re = min(re,search(a,b,l,m,2*w));
    else
    if(a>m)
    re = min(re,search(a,b,m+1,r,2*w+1));
    else
    {
        re = min(re,search(a,m,l,m,2*w));
        re = min(re,search(m+1,b,m+1,r,2*w+1));
    }
    return re;
}
int search1(int a,int b,int l,int r,int w)
{
    if(a==l&&b==r)
    {
        return ma[w];
    }
    int m = (l+r)/2,re = -1;
    if(b<=m)
    re = max(re,search1(a,b,l,m,2*w));
    else
    if(a>m)
    re = max(re,search1(a,b,m+1,r,2*w+1));
    else
    {
        re = max(re,search1(a,m,l,m,2*w));
        re = max(re,search1(m+1,b,m+1,r,2*w+1));
    }
    return re;
}
int main()
{
    int n,q,i,j,k,a,b;
    while(scanf("%d%d", &n,&q)!=EOF)
    {
        build(1,n,1);
        upset(1,n,1);
        while(q--)
        {
            scanf("%d%d", &a,&b);
            printf("%d\n",search1(a,b,1,n,1)-search(a,b,1,n,1));
        }
    }
    return 0;
}