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D. Once Again...
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.
The first line contains two space-separated integers: n, T (1 ≤ n ≤ 100, 1 ≤ T ≤ 10^7). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 300).
Print a single number — the length of a sought sequence.
4 3
3 1 4 2
5
The array given in the sample looks like that: 3, 1, 4, 2, 3, 1, 4, 2, 3, 1, 4, 2. The elements in bold form the largest non-decreasing subsequence.
题意是给出一个序列和k值，这个序列能循环k次，求其中的最长不下降子序列，k的值非常大，所以nlogn也要炸。仔细想一想，就算原数组中的所有数都进入了这个子序列中，也只有100个，剩下的肯定可以随意插入数字，那么这个数字显然是出现越多越好。那么就可以把原来的数组重复个许多(min(n, k))遍，求出其最长不下降子序列，在加上剩下的重复的序列中出现最多的数字的出现次数就是答案了。（其实我比赛时十多分钟就想到了，可惜发现以前居然把nlogn的LIS写挫了，我又抄了以前的orz）
#include<iostream> #include<stdio.h> #include<queue> #include<map> #include<algorithm> using namespace std; const int maxn = 1e4 + 10; int tt[110]; int num[maxn]; int dp[maxn]; map<int, int> mp; int main() { int n, k; while(~scanf("%d %d", &n, &k)) { int maxnum = 0, maxcnt = 0; for(int i = 1; i <= n; i++) { scanf("%d", &tt[i]); mp[tt[i]]++; if(mp[tt[i]] > maxcnt) { maxnum = tt[i]; maxcnt = mp[tt[i]]; } } int cnt = 0; for(int i = 1; i <= min(n, k); i++) { for(int j = 1; j <= n; j++) { num[++cnt] = tt[j]; } } int val = 0; //printf("%d\n", cnt); for(int i = 1; i <= cnt; i++) { int pos = upper_bound(dp, dp + val, num[i]) - dp; if(pos == val) dp[val++] = num[i]; else dp[pos] = num[i]; } if(k > n) val = val + (k - n) * maxcnt; //cout << last << " " << maxcnt << endl; printf("%d\n", val); } }
Codeforces Round #323 (Div. 2) D.Once Again...
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原文地址：http://www.cnblogs.com/lonewanderer/p/5668238.html
