codeforces C. Pearls in a Row map的应用

There are n pearls in a row. Let's enumerate them with integers from 1 to n from the left to the right. The pearl number i has the type ai.

Let's call a sequence of consecutive pearls a segment. Let's call a segment good if it contains two pearls of the same type.

Split the row of the pearls to the maximal number of good segments. Note that each pearl should appear in exactly one segment of the partition.

As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printfinstead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

Input

The first line contains integer n (1 ≤ n ≤ 3·105) — the number of pearls in a row.

The second line contains n integers ai (1 ≤ ai ≤ 109) – the type of the i-th pearl.

Output

On the first line print integer k — the maximal number of segments in a partition of the row.

Each of the next k lines should contain two integers lj, rj (1 ≤ lj ≤ rj ≤ n) — the number of the leftmost and the rightmost pearls in the j-th segment.

Note you should print the correct partition of the row of the pearls, so each pearl should be in exactly one segment and all segments should contain two pearls of the same type.

If there are several optimal solutions print any of them. You can print the segments in any order.

If there are no correct partitions of the row print the number "-1".

Sample test(s)

input

5
1 2 3 4 1

output

1
1 5

input

5
1 2 3 4 5

output

-1

input

7
1 2 1 3 1 2 1

output

2
1 3
4 7

题意：将这些算分成一段一段的，每段最多包含两个相同的数字，最多有多少段，和怎么分的段；

思路：用map记录这个数字在当前这段出现的次数，用队列存当前这段的数字，当有一个数字出现第二次时，清空队列而且保存起点和终点位置；

ＡＣ代码：

#include <bits/stdc++.h>
using namespace std;
const int N=1e5+5;
int a[5*N];
struct node
{
    int a,pos;
};
struct po
{
    int le,ri;
};
po ans[5*N];
int main()
{
    map<int,int>mp;
    queue<node>qu;
    node x;
   int n;
   int cnt=0;
   scanf("%d",&n);
   for(int i=1;i<=n;i++)
   {
       scanf("%d",&a[i]);
   }
   for(int i=1;i<=n;i++)
   {
       if(mp[a[i]]==0)
       {
           mp[a[i]]++;
           x.a=a[i];
           x.pos=i;
           qu.push(x);
       }
       else
       {
           mp[a[i]]++;
           x.a=a[i];
           x.pos=i;
           qu.push(x);
           int l,r;
           l=qu.front().pos;
           while(!qu.empty())
           {
               x=qu.front();
               mp[x.a]--;
               qu.pop();
           }
           r=x.pos;
           cnt++;
           ans[cnt].le=l;
           ans[cnt].ri=r;
       }
   }
   if(!cnt)cout<<"-1"<<endl;
   else
   {
       printf("%d
",cnt);
       for(int i=1;i<cnt;i++)
       {
           printf("%d %d
",ans[i].le,ans[i].ri);
       }
       printf("%d %d
",ans[cnt].le,n);
   }
    return 0;
}