codeforces#323(div2) D. Once Again... 求周期数列的LIS

codeforces#323(div2) D. Once Again...   求周期数列的LIS

D. Once Again...
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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.

Input

The first line contains two space-separated integers: nT (1 ≤ n ≤ 100, 1 ≤ T ≤ 107). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 300).

Output

Print a single number — the length of a sought sequence.

Sample test(s)
input
4 3
3 1 4 2
output
5
Note

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.

这题有点乱搞的意味。。思考的过程应该是这样的:

首先,先考虑T很大时,我们限定每个数只能用一次,那么LIS最多不会超过n,LIS所在的子数列长度最多也就是n*n,即每个周期都只选一个的情况,因此我们可以先求出n个周期的LIS。还剩下T-n个周期,这些周期可以插在LIS用到的那n个周期中间,这T-n个周期每个周期也要选一些数,那么选那些数呢,当然是选出现次数最多的!这样答案就是LIS(1,n*n)+(T-n)*max_cnt。

上面是T>n的情况,对于T<=n的情况,直接求LIS即可,问题就此解决!

完全不难啊。。。

#include<bits/stdc++.h>
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define MS0(a) memset(a,0,sizeof(a))

using namespace std;

typedef long long ll;
const int maxn=1000100;
const int INF=(1<<29);

int n,T;
int a[maxn];
int b[maxn],len;
int cnt[maxn];

int lis(int *a,int n)
{
    len=0;b[++len]=a[1];
    REP(i,2,n){
        if(a[i]>=b[len]) b[++len]=a[i];
        else{
            int p=upper_bound(b+1,b+len+1,a[i])-b;
            b[p]=a[i];
        }
    }
    return len;
}

int main()
{
    while(cin>>n>>T){
        MS0(cnt);
        int max_cnt=0;
        REP(i,1,n) scanf("%I64d",&a[i]),cnt[a[i]]++,max_cnt=max(max_cnt,cnt[a[i]]);
        REP(i,1,n-1){
            REP(j,1,n){
                a[i*n+j]=a[j];
            }
        }
        ll ans=0;
        if(T<=n) ans=lis(a,n*T);
        else{
            ans=lis(a,n*n)+max_cnt*(T-n);
        }
        printf("%I64d\n",ans);
    }
    return 0;
}
View Code

 

posted @ 2015-10-06 22:57  __560  阅读(385)  评论(0编辑  收藏  举报