传送门
题意
分析
我们去维护每一个节点的上一个相同数字的位置和下一个相同数字的位置,那么就可以快速处理出来每一个数距离左边的距离
我们将所有的询问进行排序,按照左端点从小到大排序,如果当前节点的下一个节点的值小于当前询问的左节点,那么当前节点的下一个节点的距离必然不会被记入答案,删除即可
我们可以用线段树去维护区间的最小值
代码
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 5e5 + 10;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {
char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}
while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
struct Node{
int l,r;
int mi;
}tr[N * 4];
int a[N],mi[N];
int pre[N],ne[N];
map<int,int> pos;
int ans[N];
int n,m;
struct qu{
int fi,se,id;
}q[N];
bool cmp(qu A,qu B){
return A.fi < B.fi;
}
void push(int u){
tr[u].mi = min(tr[u << 1].mi,tr[u << 1 | 1].mi);
}
void build(int u,int l,int r){
tr[u] = {l,r};
if(l == r) {
tr[u].mi = mi[l];
return;
}
int mid = l + r >> 1;
build(u << 1,l,mid),build(u << 1 | 1,mid + 1,r);
push(u);
}
void modify(int u,int x,int val){
if(tr[u].l == tr[u].r) {
tr[u].mi = val;
return;
}
int mid = tr[u].l + tr[u].r >> 1;
if(x <= mid) modify(u << 1,x,val);
else modify(u << 1 | 1,x,val);
push(u);
}
int query(int u,int x){
if(tr[u].r <= x) return tr[u].mi;
int mid = tr[u].l + tr[u].r >> 1;
if(x <= mid) return query(u << 1,x);
return min(query(u << 1,x),query(u << 1 | 1,x));
}
int main() {
read(n),read(m);
for(int i = 1;i <= n;i++) read(a[i]);
for(int i = 1;i <= n;i++){
pre[i] = pos[a[i]];
ne[pos[a[i]]] = i;
pos[a[i]] = i;
mi[i] = !pre[i] ? INF : i - pre[i];
}
build(1,1,n);
for(int i = 1;i <= m;i++) read(q[i].fi),read(q[i].se),q[i].id = i;
sort(q + 1,q + 1 + m,cmp);
int x = 1;
for(int i = 1;i <= m;i++){
while(x <= n && x < q[i].fi){
modify(1,ne[x++],INF);
}
int val = query(1,q[i].se);
ans[q[i].id] = val == INF ? -1 : val;
}
for(int i = 1;i <= m;i++) di(ans[i]);
return 0;
}