UVA1608-Non-boring sequences（分治）

Problem UVA1608-Non-boring sequences

Accept: 227  Submit: 2541
Time Limit: 3000 mSec

Problem 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

 

题解：典型的分治，如果找到一个数只出现一次，那么所有跨过这个数的区间都已经成立了，因此只需判断这个数左边和右边的区间的所有子区间是否成立，这样分治的感觉就很明显了，每次在所要判断的区间内找只出现一次的数字，递归判断左右，找数字的时候从左右两边开始一起找，避免最差的情况下变成O(n^2).

 

#include <bits/stdc++.h>

using namespace std;

const int maxn = 200000 + 100;

map<int, int> mmap;

int n, num[maxn];
int Next[maxn], Pre[maxn];

bool dfs(int le, int ri) {
    if (le >= ri) return true;

    int i = le, j = ri;
    int mid;
    while (i <= j) {
        if (Pre[i] < le && Next[i] > ri) {
            mid = i;
            break;
        }
        if (Pre[j] < le && Next[j] > ri) {
            mid = j;
            break;
        }
        i++, j--;
    }
    if (i > j) return false;
    return (dfs(le, mid - 1) && dfs(mid + 1, ri));
}

int main()
{
    //freopen("input.txt", "r", stdin);
    int iCase;
    scanf("%d", &iCase);
    while (iCase--) {
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) {
            scanf("%d", &num[i]);
        }

        mmap.clear();
        for (int i = 1; i <= n; i++) {
            if (!mmap.count(num[i])) {
                Pre[i] = 0;
                mmap[num[i]] = i;
            }
            else {
                int pos = mmap[num[i]];
                Pre[i] = pos;
                mmap[num[i]] = i;
            }
        }

        mmap.clear();
        for (int i = n; i >= 1; i--) {
            if (!mmap.count(num[i])) {
                Next[i] = n + 1;
                mmap[num[i]] = i;
            }
            else {
                int pos = mmap[num[i]];
                Next[i] = pos;
                mmap[num[i]] = i;
            }
        }

        bool ok = false;
        if (dfs(1, n)) ok = true;

        if (ok) printf("non-boring
");
        else printf("boring
");
    }
    return 0;
}