Codeforces #323 D. Once Again... （LIS）

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本文探讨了在给定周期序列的情况下，如何高效地求解最长不减子序列(LIS)的问题。通过预处理和利用set贪心策略，提出了一种有效的方法来解决这个问题，并给出了详细的实现步骤和代码示例。

题意:

$给出T<=10^7个周期的N<=100的序列, 求这个序列的LIS$

分析:

$因为序列a_1, a_2, ..., a_{nT} 最多有n个不同的元素, 它的任意一个不减子序列中最多有n - 1个上升元素对.$
$根据这个, 我们可以先预处理出前n个周期 (a_1, ..., a_{n^2}), 以及后n个周期(a_{nT - n + 1}, ..., a_{nT})的LIS.$
$f[i]:=以a[i]结尾的前n个周期的LIS的最大长度$
$g[i]:=以a[i]开头的后n个周期的LIS的最大长度$
$ans = max\{f[i] + (T-2n)count(a_i)+max\{g[j], a_j >= a_i\cap j \in[nT-n+1,nT]\}, i \in[1, n^2]\}$
$ps: count(a_i) = cnt \ of \ a_i \ in\ a_1, ..., a_n$
$怎么来求这个呢, 嘛, 这一个很明显的set贪心辣, sort一下维护满足条件的g[j]最大值, 然后不断更新ans即可$

代码:

//
//  Created by TaoSama on 2015-10-04
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, T, a[N], f[N], g[N], h[N], cnt[305];
//f: end i g: start i

int r[N];
bool cmp(int x, int y) {
    return a[x] > a[y];
}

int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
//  freopen("out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    while(scanf("%d%d", &n, &T) == 2) {
        memset(cnt, 0, sizeof cnt);
        for(int i = 1; i <= n; ++i) {
            scanf("%d", a + i);
            ++cnt[a[i]];
        }
        int t = min(T, n << 1);
        for(int i = 1; i < t; ++i)
            for(int j = 1; j <= n; ++j)
                a[j + i * n] = a[j];

        int ans = 0;
        if(T <= n << 1) {
            memset(h, 0x3f, sizeof h);
            for(int i = 1; i <= n * t; ++i) {
                int k = upper_bound(h + 1, h + n * t + 1, a[i]) - h;
                ans = max(ans, k);
                h[k] = a[i];
            }
        } else {
            memset(h, 0x3f, sizeof h);
            for(int i = 1; i <= n * n; ++i) {
                int k = upper_bound(h + 1, h + n * n + 1, a[i]) - h;
                f[i] = k;
                h[k] = a[i];
            }

            memset(h, 0x3f, sizeof h);
            for(int i = n * n; i; --i) {
                int k = upper_bound(h + 1, h + n * n + 1, -a[i]) - h;
                g[i] = k;
                h[k] = -a[i];
            }

            for(int i = 1; i <= n * n; ++i) r[i] = i;
            sort(r + 1, r + n * n + 1, cmp);
            set<int> s;
            for(int i = 1, j = 1; i <= n * n; ++i) {
                int x = r[i];
                while(j <= n * n && a[r[j]] >= a[r[i]]) s.insert(g[r[j++]]);
                int tmp = f[x] + (T - 2 * n) * cnt[a[x]] + (*s.rbegin());
                ans = max(ans, tmp);
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}

UPD: $其实暴力n段, 将出现最多的插入中间也能ac, 虽然正确性不是那么显然$
代码:

//
//  Created by TaoSama on 2015-10-04
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e6 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, T, a[N], g[N];
int cnt[305];

int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
//  freopen("out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    while(scanf("%d%d", &n, &T) == 2) {
        memset(cnt, 0, sizeof cnt);
        int maxv = 0;
        for(int i = 1; i <= n; ++i) {
            scanf("%d", a + i);
            maxv = max(maxv, ++cnt[a[i]]);
        }
        int t = min(T, n);
        for(int i = 1; i < t; ++i)
            for(int j = 1; j <= n; ++j)
                a[j + i * n] = a[j];

        memset(g, 0x3f, sizeof g);
        int ans = 0;
        for(int i = 1; i <= n * t; ++i) {
            int k = upper_bound(g + 1, g + n * t + 1, a[i]) - g;
            ans = max(ans, k);
            g[k] = a[i];
        }

        ans += maxv * (T - t);
        printf("%d\n", ans);
    }
    return 0;
}