Codeforces VK Cup 2015

D. Closest Equals

Time Limit: 1 Sec  Memory Limit: 256 MB

题目连接

http://www.lydsy.com/JudgeOnline/problem.php?id=3224

Description

You are given sequence a₁, a₂, ..., a_n and m queries l_j, r_j (1 ≤ l_j ≤ r_j ≤ n). For each query you need to print the minimum distance between such pair of elements a_x and a_y (x ≠ y), that:

• both indexes of the elements lie within range [l_j, r_j], that is, l_j ≤ x, y ≤ r_j;
• the values of the elements are equal, that is a_x = a_y.

The text above understands distance as |x - y|.

Input

The first line of the input contains a pair of integers n, m (1 ≤ n, m ≤ 5·10⁵) — the length of the sequence and the number of queries, correspondingly.

The second line contains the sequence of integers a₁, a₂, ..., a_n ( - 10⁹ ≤ a_i ≤ 10⁹).

Next m lines contain the queries, one per line. Each query is given by a pair of numbers l_j, r_j (1 ≤ l_j ≤ r_j ≤ n) — the indexes of the query range limits.

Output

Print m integers — the answers to each query. If there is no valid match for some query, please print -1 as an answer to this query.

Sample Input

5 3
1 1 2 3 2
1 5
2 4
3 5

Sample Output

1
-1
2

HINT

题意

 查询区间相同数的最小距离

题解：

用一个map记录前面的位置，然后离线搞一搞
用心去体会，我也不好说……

单点更新，区间查询最小值

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 500001
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
/*

inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
*/
//**************************************************************************************
inline ll read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}

struct node{
    int l,r,v;
}a[maxn*4];
struct ques
{
    int l,r,an,v;
}qu[maxn];
void build(int x,int l,int r)
{
    a[x].l=l,a[x].r=r;
    a[x].v=maxn;
    if(l==r)
        return;
    int mid=(l+r)>>1;
    build(x<<1,l,mid);
    build(x<<1|1,mid+1,r);
}
void pushup(int x)
{
    a[x].v=min(a[x<<1].v,a[x<<1|1].v);
}
void update(int x,int pos,int val)
{
    if(a[x].l==a[x].r)
    {
        a[x].v=val;
        return;
    }
    int mid=(a[x].l+a[x].r)>>1;
    if(pos<=mid)
        update(x<<1,pos,val);
    else
        update(x<<1|1,pos,val);
    pushup(x);
}
int mi;
void query(int x,int l,int r)
{
    if(l<=a[x].r&&r>=a[x].r)
    {
        mi=min(mi,a[x].v);
        return;
    }
    int mid=(a[x].l+a[x].r)>>1;
    if(l<=mid)
        query(x<<1,l,r);
    if(r>mid)
        query(x<<1|1,l,r);
}
map<int,int>mp;
int d[maxn];
bool cmp(ques a,ques b)
{
    return a.l>b.l;
}
int ans[maxn];
int main()
{
    int n=read(),m=read();
    build(1,1,n);
    for(int i=1;i<=n;i++)
        d[i]=read();
    for(int i=1;i<=m;i++)
    {
        qu[i].l=read(),qu[i].r=read();
        qu[i].v=i;
    }
    sort(qu+1,qu+1+m,cmp);
    int t=1;
    for(int i=n;i;i--)
    {
        if(mp[d[i]])
        {
            update(1,mp[d[i]],mp[d[i]]-i);
        }
        mp[d[i]]=i;
        while(qu[t].l==i)
        {
            mi=maxn;
            query(1,qu[t].l,qu[t].r);
            if(mi==maxn)
                mi=-1;
            ans[qu[t].v]=mi;
            t++;
        }
    }
    for(int i=1;i<=m;i++)
    {
        if(ans[i]<0)
            printf("-1
");
        else
            P(ans[i]);
    }
}