Codeforces 582B Once Again... 【LIS变形】

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最新推荐文章于 2024-07-22 14:40:37 发布

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分类专栏： DP 之 LIS codeforces

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本文介绍了解决Codeforces582B问题的方法，该问题是寻找给定序列重复多次后形成的数组中最长的非递减子序列。通过分析，提出了将序列分为两部分并分别计算最长非递减子序列长度的策略。

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题目链接：Codeforces 582B Once Again…

B. Once Again…
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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: n, T (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.

Examples
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

题意：给定n个元素的序列，现在将它重复T次。问你可以得到的LIS。

思路：当n个元素各不相同时，在有n+1段的情况下会导致有一段的贡献必定为1。这样的话就好办了，我们取前n段为一个序列A[]，后n段为一个序列B[]。
dp1[i]记录以A[i]结尾的LIS，dp2[i]记录以B[i]开头的LIS。我们暴力起点i、终点j，中间序列取合法的A[k](A[i] <= A[k] <= A[j])且A[k]数量最多。扫描一遍就好了。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <map>
#include <set>
#include <string>
#include <queue>
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int MAXN = 2*1e4 + 10;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
int g[MAXN], a[MAXN];
int dp1[MAXN], dp2[MAXN];
int cnt[400];
int main()
{
    int n, T;
    while(scanf("%d%d", &n, &T) != EOF) {
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]); g[i] = INF;
        }
        int ans = 0;
        if(T <= 2*n) {
            for(int i = n+1; i <= T * n; i++) {
                a[i] = a[i % n == 0 ? n : i % n];
                g[i] = INF;
            }
            for(int i = 1; i <= T * n; i++) {
                int k = upper_bound(g+1, g+T*n+1, a[i]) - g;
                dp1[i] = k;
                g[k] = min(g[k], a[i]);
                ans = max(ans, dp1[i]);
            }
            printf("%d\n", ans);
            continue;
        }
        for(int i = n+1; i <= n * n; i++) {
            a[i] = a[i % n == 0 ? n : i % n];
            g[i] = INF;
        }
        for(int i = 1; i <= n * n; i++) {
            int k = upper_bound(g+1, g+n*n+1, a[i]) - g;
            dp1[i] = k;
            g[k] = min(g[k], a[i]);
        }
        CLR(g, INF);
        for(int i = n * n; i >= 1; i--) {
            int k = upper_bound(g+1, g+n*n+1, -a[i]) - g;
            dp2[i] = k;
            g[k] = min(g[k], -a[i]);
        }
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= n; j++) {
                if(a[i] <= a[j]) {
                    int Max = 0; CLR(cnt, 0);
                    for(int k = 1; k <= n; k++) {
                        if(a[k] >= a[i] && a[k] <= a[j]) {
                            cnt[a[k]]++;
                        }
                        Max = max(Max, cnt[a[k]]);
                    }
                    ans = max(ans, dp1[(n - 1) * n + i] + dp2[j] + Max * (T - 2*n));
                }
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}