http://codeforces.com/contest/583/problem/D
原题:You are given an array of positive integers a1, a2, ..., an × T of length n × T. We know that for any i > n it is true that ai = ai - n. Find the length of the longest non-decreasing sequence of the given array.
题目大意:有长度为n的数组a(n <= 100),其中a[i] <= 300,这个a数组可以重复T次,问他的最长上升子序列是多少?
思路:我们可以发现,这个数组如果要全部都算上的,那么在t<=n的情况下,他的最长上升子序列一定会遍历一次a数组。所以我们就只需要把原来的数组扩大n倍,然后求他的LIS。
这样以后我们发现,后面的重复的次数一定是原来数组里面出现次数(假定重复次数k为最多)最多的数值,所以ans = Lis的长度 + k * T - min(n, T);
复杂度 O(n*n*logn)
//看看会不会爆int!数组会不会少了一维! //取物问题一定要小心先手胜利的条件 #include <bits/stdc++.h> using namespace std; #pragma comment(linker,"/STACK:102400000,102400000") #define LL long long #define ALL(a) a.begin(), a.end() #define pb push_back #define mk make_pair #define fi first #define se second #define haha printf("haha ") const int maxn = 100 + 5; int a[maxn * maxn]; int n, T; vector<int> ve; int solve(int t){ int len = t * n; for (int i = 1; i < t; i++){ for (int j = 1; j <= n; j++){ a[n * i + j] = a[j]; } } for (int i = 1; i <= n * t; i++){ int pos = upper_bound(ve.begin(), ve.end(), a[i]) - ve.begin(); if (pos == ve.size()) ve.push_back(a[i]); else ve[pos] = a[i]; } return ve.size(); } int cnt[maxn * maxn]; int main(){ cin >> n >> T; int maxval = 0, k = 0; for (int i = 1; i <= n; i++){ scanf("%d", a + i); cnt[a[i]]++; k = max(cnt[a[i]], k); } int ans = solve(min(n, T)); //printf("ans = %d k = %d T - min(n, T) = %d ", ans, k, T - min(n, T)); printf("%d ", ans + k * (T - min(n, T))); return 0; }