UVA - 1608

We were afraid of making this problem statement too boring, so we decided to keep it short. A sequence
is called non-boring if its every connected subsequence contains a unique element, i.e. an element such
that no other element of that subsequence has the same value.
Given a sequence of integers, decide whether it is non-boring.

Input

The first line of the input contains the number of test cases T. The descriptions of the test cases follow:
Each test case starts with an integer n (1 ≤ n ≤ 200000) denoting the length of the sequence. In
the next line the n elements of the sequence follow, separated with single spaces. The elements are
non-negative integers less than 109.

Output

Print the answers to the test cases in the order in which they appear in the input. For each test case
print a single line containing the word ‘non-boring’ or ‘boring’.

Sample Input

4
5
1 2 3 4 5
5
1 1 1 1 1
5
1 2 3 2 1
5
1 1 2 1 1

Sample Output

non-boring
boring
non-boring
boring

一组数列，如果其中有一个数字只出现过一次，则可以判定它是non-boring的。
先找出母序列中唯一的数，它在母序列中是唯一的，那么在子序列中也是唯一的，则所有跨过它的子序列也都是non-boring的，不用判断，所以可以递归查找数字两边的子序列，直到查到长度为1则可以返回true。
那么如何判断这个数在序列中是唯一的？可以在读入的时候记录这个数字上一次出现和下一次出现的位置，判断是否在数列的范围内即可。

#include <stdio.h>
#include <map>
using namespace std;

int l[200015],r[200015],a[200015];

bool sss(int x,int y){
    if(x>=y){
        return true;
    }
    for(int i=0;i<=(y-x)/2;i++){
        if(l[x+i]<x && r[x+i]>y){
            return sss(x,x+i-1) && sss(x+i+1,y);
        }
        if(l[y-i]<x&&r[y-i]>y){
            return sss(x,y-i-1) && sss(y-i+1,y);
        }
    }
    return false;
}

int main()
{
    int T,n,i,j,k;
    scanf("%d",&T);
    while(T--){
        map<int,int> m;
        scanf("%d",&n);
        for(i=0;i<n;i++){
            scanf("%d",&a[i]);
        }
        for(i=0;i<n;i++){
            if(!m.count(a[i])){
                l[i]=-1;
            }else{
                l[i]=m[a[i]];
            }
            m[a[i]]=i;
        }
        m.clear();
        for(i=n-1;i>=0;i--){
            if(!m.count(a[i])){
                r[i]=n;
            }else{
                r[i]=m[a[i]];
            }
            m[a[i]]=i;
        }
        m.clear();

        if(sss(0,n-1)){
            printf("non-boring\n");
        }else{
            printf("boring\n");
        }
    }
    return 0;
}