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 secondsmemory 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<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
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<int> 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<Node> 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<Node> 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-----------------------------------------------------------------------
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