VK Cup 2015 - Qualification Round 1 D. Closest Equals 离线+线段树

题目链接：

http://codeforces.com/problemset/problem/522/D

D. Closest Equals

time limit per test3 seconds
memory limit per test256 megabytes #### 问题描述 > You are given sequence a1, a2, ..., an and m queries lj, rj (1 ≤ lj ≤ rj ≤ n). For each query you need to print the minimum distance between such pair of elements ax and ay (x ≠ y), that: > > both indexes of the elements lie within range [lj, rj], that is, lj ≤ x, y ≤ rj; > the values of the elements are equal, that is ax = ay. > The text above understands distance as |x - y|.

输入

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, ..., an ( - 109 ≤ ai ≤ 109).

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

输出

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.

样例输入

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

样例输出

1
-1
2

题意

求区间内相邻最近的两个相同的数的距离。

题解

线段树，先处理出所有的相同的数的相邻的间隔区间，把这些区间（之后称为事件）按右端点排序，对于所有的查询区间也同样排序，然后一边扫查询，一边插入事件，事件按左端点插入，值为区间大小。

代码

include

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include

include

include

include

include

include

include

include

include

include

include

include

using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=5e5+10;

int minv[maxn<<2];

int ql,qr,qmin;
void query(int o,int l,int r){
if(ql<=l&&r<=qr){
qmin=min(qmin,minv[o]);
}else{
if(ql<=mid) query(lson,l,mid);
if(qr>mid) query(rson,mid+1,r);
}
}

int _p,uv;
void update(int o,int l,int r){
if(l==r){
minv[o]=uv;
}else{
if(_p<=mid) update(lson,l,mid);
else update(rson,mid+1,r);
minv[o]=min(minv[lson],minv[rson]);
}
}

struct Node{
int l,r,v;
Node(int l,int r,int v):l(l),r(r),v(v){}
};

bool cmp(const Node& n1,const Node& n2){
return n1.r<n2.r;
}

map<int,int> mp;
int ans[maxn],arr[maxn],n,m;

void init(){
rep(i,0,maxn<<2) minv[i]=INF;
}

int main() {
scf("%d%d",&n,&m);
init();
vector lis;
for(int i=1;i<=n;i++){
scf("%d",&arr[i]);
if(mp[arr[i]]){
lis.pb(Node(mp[arr[i]],i,i-mp[arr[i]]));
}
mp[arr[i]]=i;
}

// rep(i,0,lis.sz()) prf("(%d,%d)\n",lis[i].l,lis[i].r);

vector que;
for(int i=0;i<m;i++){
int l,r;
scf("%d%d",&l,&r);
que.pb(Node(l,r,i));
}

sort(all(lis),cmp);
sort(all(que),cmp);

int p=0;
for(int i=0;i<que.sz();i++){
while(p<lis.sz()&&lis[p].r<=que[i].r){

_p=lis[p].l; uv=lis[p].v;
update(1,1,n);

p++;
}

ql=que[i].l,qr=que[i].r,qmin=INF;
query(1,1,n);

ans[que[i].v]=qmin>=INF?-1:qmin;
}

for(int i=0;i<m;i++) prf("%d\n",ans[i]);

return 0;
}

//end-----------------------------------------------------------------------