POJ3264 【RMQ基础题—ST-线段树】

ST算法Code：

//#include<bits/stdc++.h>
#include<cstdio>
#include<math.h>
#include<iostream>
#include<queue>
#include<algorithm>
#include<string.h>
using namespace std;
typedef long long LL;

const int N=5e4+10;

int n,q;
int a[N];
int f1[N][30];
int f2[N][30];

void ST()
{
    for(int i=1;i<=n;i++)
        f1[i][0]=f2[i][0]=a[i];
    int nlog=(int)(log(double(n))/log(2.0));
    for(int j=1;j<=nlog;j++)
    {
        for(int  i=1;i<=n;i++)
        {
            if(i+(1<<j)-1<=n)
            {
                f1[i][j]=min(f1[i][j-1],f1[i+(1<<(j-1))][j-1]);
                f2[i][j]=max(f2[i][j-1],f2[i+(1<<(j-1))][j-1]);
            }
        }
    }
}
int RMQ(int l,int r)
{
    int nlog=(int)(log(double(r-l+1))/log(2.0));
    int mi=min(f1[l][nlog],f1[r-(1<<nlog)+1][nlog]);
    int ma=max(f2[l][nlog],f2[r-(1<<nlog)+1][nlog]);
    return ma-mi;
}

int main()
{
    int l,r;
    while(scanf("%d%d",&n,&q)!=EOF)
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        ST();
        while(q--)
        {
            scanf("%d%d",&l,&r);
            printf("%d\n",RMQ(l,r));
        }
    }
    return 0;
}

线段树Code：

#include<cstdio>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;

const int N=50007;
struct st{
	int left,right;
	int mina;
	int maxa;
};
st q[N*4];
int n,m;

void build(int num,int L,int R)
{
	q[num].left=L;
	q[num].right=R;
	if(L==R)
	{
		scanf("%d",&q[num].maxa);
		q[num].mina=q[num].maxa;
		return;
	}
	build(2*num,L,(L+R)/2);
	build(2*num+1,(L+R)/2+1,R);
	q[num].mina=min(q[2*num].mina,q[2*num+1].mina);
	q[num].maxa=max(q[2*num].maxa,q[2*num+1].maxa);
}
int ans1;
int ans2;
int get_max(int num,int s,int t)
{
	if(q[num].left>=s&&q[num].right<=t)
		return q[num].maxa;
	int mid=(q[num].left+q[num].right)/2;
	if(mid>=t)
		return get_max(2*num,s,t);
	else if(mid<s)
		return get_max(2*num+1,s,t);
	else{
		return max(get_max(2*num,s,mid),get_max(2*num+1,mid+1,t));
	}
}

int get_min(int num,int s,int t)
{
	if(q[num].left>=s&&q[num].right<=t)
		return q[num].mina;
	int mid=(q[num].left+q[num].right)/2;
	if(mid>=t)
		return get_min(2*num,s,t);
	else if(mid<s)
		return get_min(2*num+1,s,t);
	else{
		return min(get_min(2*num,s,mid),get_min(2*num+1,mid+1,t));
	}
}

int main()
{
	scanf("%d%d",&n,&m);
	build(1,1,n);
	while(m--)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		printf("%d\n",get_max(1,a,b)-get_min(1,a,b));
	}
	return 0;
}