D. Closest Equals
Time Limit: 1 Sec Memory Limit: 256 MB
题目连接
https://codeforces.com/problemset/problem/522/DDescription
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|.
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, ..., 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.
Output
Sample Input
1 1 2 3 2
1 5
2 4
3 5
Sample Output
-1
2
HINT
题意
查询区间相同数的最小距离
题解：
首先对于每个数x我们只需要知道距离最近的x的位置即可，想一想为什么。
具体做法：
１．先离散，然后搞一个nex[]，nex[i]表示位置i右边和它权值相等且最近的位置。
２．这样对于每个询问(l,r),只有当i和nex[i]都在(l,r)内才能计算贡献。
３．看起来很麻烦其实我们可以离线做，将询问按左端点从小到大排序，
先将(1,n)赋值成INF(一个很大的数),
当扫到区间(l,r)时，对于i(l<=i<=r),我们将nex[i]位置赋值为nex[i]-i，然后直接求(l,r)的最小值即可。
注意当i<l后记得将nex[i]重新赋值为INF。
4.推荐用树状数组做一下《HH的项链》，思想有异曲同工之妙。
Debug：
读入优化忘记处理负数，一直wa在19个点，真是悲伤，数据太水了吧，前面的负数是怎过的？
代码：
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define N 1000050 4 #define INF 12345678 5 #define LL long long 6 LL n,m,cnt,kth[N],a[N],nex[N],last[N],ans[N]; 7 struct Query 8 { 9 LL l,r,id; 10 bool operator <(const Query&b)const 11 {return l<b.l;} 12 }q[N]; 13 struct Tree{LL l,r,min;}tr[N<<2]; 14 template<typename T>void read(T&x) 15 { 16 LL k=0;char c=getchar(); 17 x=0; 18 while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar(); 19 if(c==EOF)exit(0); 20 while(isdigit(c))x=x*10+c-'0',c=getchar(); 21 } 22 void bt(LL x,LL l,LL r) 23 { 24 tr[x]=Tree{l,r,INF}; 25 if(l==r)return; 26 LL mid=(l+r)>>1; 27 bt(x<<1,l,mid); 28 bt(x<<1|1,mid+1,r); 29 } 30 void update(LL x,LL p,LL tt) 31 { 32 if (p<=tr[x].l&&tr[x].r<=p) 33 { 34 tr[x].min=tt; 35 return; 36 } 37 LL mid=(tr[x].l+tr[x].r)>>1; 38 if(p<=mid)update(x<<1,p,tt); 39 if (mid<p)update(x<<1|1,p,tt); 40 tr[x].min=min(tr[x<<1].min,tr[x<<1|1].min); 41 } 42 LL query(LL x,LL l,LL r) 43 { 44 if(l<=tr[x].l&&tr[x].r<=r) 45 return tr[x].min; 46 LL mid=(tr[x].l+tr[x].r)>>1,ans=INF; 47 if(l<=mid)ans=query(x<<1,l,r); 48 if(mid<r)ans=min(ans,query(x<<1|1,l,r)); 49 return ans; 50 } 51 int main() 52 { 53 #ifndef ONLINE_JUDGE 54 freopen("aa.in","r",stdin); 55 #endif 56 read(n); read(m); 57 bt(1,1,n); 58 for(LL i=1;i<=n;i++)read(a[i]),kth[i]=a[i]; 59 sort(kth+1,kth+n+1); 60 cnt=unique(kth+1,kth+n+1)-kth-1; 61 for(LL i=1;i<=n;i++)a[i]=lower_bound(kth+1,kth+cnt+1,a[i])-kth; 62 for(LL i=1;i<=n;i++) 63 { 64 if (last[a[i]])nex[last[a[i]]]=i; 65 last[a[i]]=i; 66 } 67 for(LL i=1;i<=m;i++) 68 { 69 q[i].id=i; 70 read(q[i].l); read(q[i].r); 71 } 72 sort(q+1,q+m+1); 73 LL l=q[1].l,r=q[1].l-1; 74 for(LL i=1;i<=m;i++) 75 { 76 for(LL j=l;j<=q[i].l-1;j++) 77 if (nex[j]>0)update(1,nex[j],INF); 78 for(LL j=r+1;j<=q[i].r;j++) 79 if (nex[j]>0)update(1,nex[j],nex[j]-j); 80 ans[q[i].id]=query(1,q[i].l,q[i].r); 81 if(ans[q[i].id]==INF)ans[q[i].id]=-1; 82 l=q[i].l; 83 r=max(r,q[i].r); 84 } 85 for(LL i=1;i<=m;i++) 86 printf("%lld ",ans[i]); 87 } 88