POJ 3264 Balanced Lineup（st或者线段树）

标签：rmq问题st算法   线段树单点更新   

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#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define maxn 50005
int n,q;
int dp[maxn][30],d[maxn][30],a[maxn];
void rmq(){
   for(int i=1;i<=n;i++)dp[i][0]=a[i],d[i][0]=a[i];
   for(int j=1;(1<<j)<=n;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]);
        d[i][j]=max(d[i][j-1],d[i+(1<<(j-1))][j-1]);
    }
   }
}
int query1(int l,int r){
    int k=0;
    while((1<<(k+1))<=r-l+1)k++;
    return min(dp[l][k],dp[r-(1<<k)+1][k]);
}
int query2(int l,int r){
    int k=0;
    //int k=floor(log(R-L+1.0)/log(2.0));
    while((1<<(k+1))<=r-l+1)k++;
    return max(d[l][k],d[r-(1<<k)+1][k]);
}
int main()
{
    int u,v;
    //freopen("in.txt","r",stdin);
    while(~scanf("%d%d",&n,&q)){
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
        }
        rmq();
        for(int i=0;i<q;i++){
            scanf("%d%d",&u,&v);
            printf("%d\n",query2(u,v)-query1(u,v));
        }

    }
}

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define maxn 50005
#define inf 0x3f3f3f3f
int n,q;
int a[maxn],ll[maxn<<1],rr[maxn<<1],mi[maxn<<1],ma[maxn<<1];
inline void pushup(int i){
    ma[i]=max(ma[i<<1],ma[i<<1|1]);
    mi[i]=min(mi[i<<1],mi[i<<1|1]);
}
void build(int l,int r,int i){
    ll[i]=l;
    rr[i]=r;
    if(l==r){
        mi[i]=a[l];
        ma[i]=a[l];
        return ;
    }
    int m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;
    build(l,m,ls);
    build(m+1,r,rs);
    pushup(i);
}
int query1(int l,int r,int i){
    if(l<=ll[i]&&rr[i]<=r){
        return mi[i];
    }
    int m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;
    int ans=inf;
    if(l<=m)ans=min(ans,query1(l,r,ls));
    if(m<r)ans=min(ans,query1(l,r,rs));
    return ans;
}
int query2(int l,int r,int i){
    if(l<=ll[i]&&rr[i]<=r){
        return ma[i];
    }
    int m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;
    int ans=0;
    if(l<=m)ans=max(ans,query2(l,r,ls));
    if(m<r)ans=max(ans,query2(l,r,rs));
    return ans;
}
int main()
{
    int u,v;
    //freopen("in.txt","r",stdin);
    while(~scanf("%d%d",&n,&q)){
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
        }
        build(1,n,1);
        for(int i=0;i<q;i++){
            scanf("%d%d",&u,&v);
            printf("%d\n",query2(u,v,1)-query1(u,v,1));
        }

    }
}

POJ 3264 Balanced Lineup（st或者线段树）

标签：rmq问题st算法   线段树单点更新   

原文地址：http://blog.csdn.net/u013497977/article/details/47364603