1 /**\
2 https://codeforces.com/contest/522/problem/D
3 离线查询, 按照右端点小的排前面,
4 使用p[x] 记录a[x]出现相同值的前一个位置, p[x]当作线段树的坐标, x - p[x]为值,
5 olgN进行加点, 然后查询
6 \**/
7 #include <bits/stdc++.h>
8 using namespace std;
9 #define fi first
10 #define se second
11
12 #define go continue
13 #define int long long
14 #define PII pair<int, int>
15 #define sf(x) scanf("%lld",&x)
16 #define ytz int _; sf(_); while(_--)
17
18 #define RE(i, a) for(int i = 1; i <= a; ++i)
19 #define debug(a) cout << #a << " = " << a <<endl;
20
21 const int N = 5e5 + 10;
22 const int inf = 0x3f3f3f3f;
23
24 int n, m;
25 int ok[N], p[N];
26 unordered_map<int, int> vis;
27
28 struct Query
29 {
30 int l, r, id;
31 bool operator < (const Query& t) const
32 {
33 return r < t.r;
34 }
35 } q[N];
36
37 struct node
38 {
39 int l, r, mn;
40 } t[N << 2];
41 int mp[N]; //快速单点修改, 记录底层叶子节点的root值
42
43 inline void build(int root, int l, int r)
44 {
45 t[root].l = l, t[root].r = r;
46 t[root].mn = inf;
47 if(l == r)
48 {
49 mp[l] = root;
50 return;
51 }
52 int ch = root << 1;
53 int mid = (l + r) >> 1;
54 build(ch, l, mid);
55 build(ch + 1, mid + 1, r);
56 }
57
58 inline void update(int x, int y)
59 {
60 x = mp[x];
61 t[x].mn = (t[x].mn, y);
62 while(x >>= 1)
63 {
64 int ch = x << 1;
65 t[x].mn = min(t[ch].mn, t[ch + 1].mn);
66 }
67 }
68
69 inline int query(int root, int l, int r)
70 {
71 if(t[root].l >= l && t[root].r <= r)
72 {
73 return t[root].mn;
74 }
75 int ch = root << 1;
76 int mid = (t[root].l + t[root]. r) >> 1;
77 if(r <= mid) return query(ch, l, r);
78 else if(l > mid) query(ch + 1, l, r);
79 else return min(query(ch, l, mid), query(ch + 1, mid + 1, r));
80 }
81
82 signed main()
83 {
84 sf(n), sf(m);
85 build(1, 1, n);
86
87 int x;
88 RE(i, n)
89 {
90 sf(x);
91
92 if(vis[x]) p[i] = vis[x], vis[x] = i;
93 else vis[x] = i, p[i] = 0;
94 }
95 RE(i, m)
96 {
97 sf(q[i].l), sf(q[i].r);
98 q[i].id = i;
99 }
100 sort(q + 1, q + 1 + m);
101 int id = 1;
102 RE(i, m)
103 {
104 while(id <= q[i].r)
105 {
106 if(p[id]) update(p[id], id - p[id]);
107 id++;
108 }
109 int tmp = query(1, q[i].l, id - 1);
110 if(tmp >= inf) ok[q[i].id] = -1;
111 else ok[q[i].id] = tmp;
112 }
113 RE(i, m) printf("%lld\n", ok[i]);
114 return 0;
115 }
