UVA 1608 Non-boring sequence
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题目大意就是判断一个序列中,对于任意子序列是否存在重复元素,如果是则序列是无聊的,否则是不无聊的。
思路
分治,中途相遇法
紫书上的思路已经很清晰了,首先是找到一个当前序列中不重复的元素,递归判断被它分隔的左序列和右序列是否满足题意。
直到找到只有一个元素的序列为止。如果在扫描一个序列是从左到右扫描,最坏情况是在另一端,则 时间复杂度是O(n^2) 如
果用中途相遇法,对于每一个序列最坏情况是在中间,满足T(n) = 2 * T(n / 2) + O(n) 即时间复杂度为O(nlogn)
开始预处理一下相邻最近元素的位置即可用O(1)时间判断该元素是否在当前序列是否是独一无二的.
#include <bits/stdc++.h>using namespace std;const int maxn = 200000 + 10;int a[maxn], n;int L[maxn], R[maxn];void init(){ for(int i = 1; i <= n; ++i){ L[i] = -1; R[i] = n + 1; } map<int, int> last; for(int i = 1; i <= n; ++i){ if(!last.count(a[i])){ last[a[i]] = i; }else{ L[i] = last[a[i]]; last[a[i]] = i; } } last.clear(); for(int i = n; i >= 1; --i){ if(!last.count(a[i])){ last[a[i]] = i; }else{ R[i] = last[a[i]]; last[a[i]] = i; } }}bool solve(int x, int y){ if(y <= x) return true; int o = ((y - x) % 2 == 1); for(int k = 0; k <= (y - x + o) / 2; ++k){ if(L[x + k] < x && R[x + k] > y){ if(solve(x, x + k - 1) && solve(x + k + 1, y)) return true; }else if(L[y - k] < x && R[y - k] > y){ if(solve(x, y - k - 1) && solve(y - k + 1, y)) return true; } } return false;}int main(){ int T; scanf("%d", &T); while(T --){ scanf("%d", &n); for(int i = 1; i <= n; ++i) scanf("%d", a + i); init(); if(solve(1, n)){ printf("non-boring\n"); }else{ printf("boring\n"); } } return 0;}
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