UVA

Description

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

题目大意：

给你一个数字序列，如果它的每一个连续子串都包含不重复的数字，那么则称之为non-boring。

分析：

1.如果整个序列中存在只出现过一次的数字，那么所有包含这个数字的子串都是满足要求的，因此我们只需要检验该数字的左边区间和右边区间是否满足条件；

2.如果我们预记录下每个元素左边和右边最近的相同元素的位置，那么我们就可以O(1)的时间里判断该元素在区间内是否唯一存在；

3.如果从左往右找，最坏的情况是需要遍历整个区间。采用从两边向中间找。

#include<cstdio>#include<cstring>#include<string>#include<iostream>#include<map>#include<queue>using namespace std;typedef long long LL;const int maxn = 200002;int b[maxn][2];//左右最近的相同的数的下标int a[maxn];map<int, int >mp;//该元素最后一次出现的下标bool unique(int i, int l, int r) {    return b[i][0]<l&&b[i][1]>r;}bool check(int l, int r) {    if (l >= r) return true;    for (int d = 0; l + d <= r - d; d++) {        if (unique(l + d, l, r))            return check(l, l + d - 1) && check(l + d + 1, r);        if(unique(r-d,l,r))            return check(l, r - d - 1) && check(r - d + 1, r);        if (2 * d == r - l)//如果始终没有找到unique的元素，那么return false            return false;    }    return false;}int main() {#ifdef _DEBUG    freopen("liu.in", "r", stdin);#endif    int T; scanf("%d", &T);    while (T--) {        int n; scanf("%d", &n);        //{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{        //利用map和数组记录每个元素最近的相同元素的下标        mp.clear();        for (int i = 0; i < n; i++) {            scanf("%d", a + i);            if (!mp.count(a[i]))                b[i][0] = -1;            else                b[i][0] = mp[a[i]];            mp[a[i]] = i;        }        mp.clear();        for (int i = n - 1; i >= 0; i--) {            if (!mp.count(a[i]))                b[i][1] = n;            else                b[i][1] = mp[a[i]];            mp[a[i]] = i;        }        //}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}        puts(check(0, n - 1) ? "non-boring" : "boring");    }    return 0;}