Problem Solving►UVa

UVa 11456 - Trainsorting

Contents

  1. 1. Problem
  2. 2. Solution
  3. 3. Code

Problem

題目網址
中文網址

Solution

由於可以將火車選擇皆在前方或後方，所以將原本的數列前接上顛倒的。

1 2 3 4 =>
4 3 2 1 1 2 3 4

以 2 來說它可以是在 1 前面或後面。

接著算 LIS 即可。

Code

$O(NlogL)$

UVa 11456
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#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

int main()
{
    int Case, train[4000];
    vector<int> lis;
    scanf("%d", &Case);
    while (Case--)
    {
        int n;
        scanf("%d", &n);
        for (int i = 0; i < n; ++i)
        {
            scanf("%d", &train[i + n]);
            train[n - i - 1] = train[i + n];
        }

        if (n)
            lis.push_back(train[0]);

        for (int i = 1; i < (n << 1); ++i)
        {
            if (train[i] > lis.back())
                lis.push_back(train[i]);
            else
                *(lower_bound(lis.begin(), lis.end(), train[i])) = train[i];
        }

        printf("%d\n", lis.size());
        lis.clear();
    }

    return 0;
}

$O(N^2)$

UVa 11456
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#include <cstdio>
#include <cstring>
#define MAX(a, b) ((a) > (b) ? (a) : (b))

int main()
{
    int Case, train[4000], len[4000];
    scanf("%d", &Case);
    while (Case--)
    {
        int n;
        scanf("%d", &n);
        */
        for (int i = 0; i < n; ++i)
        {
            scanf("%d", &train[i + n]);
            train[n - i - 1] = train[i + n];
        }

        n <<= 1;
        int ans = 0;

        for (int i = 0; i < n; ++i)
        {
            len[i] = 1;
            for (int j = 0; j < i; ++j)
            {
                if (train[i] > train[j])
                    len[i] = MAX(len[i], len[j] + 1);
            }
            ans = MAX(ans, len[i]);
        }

        printf("%d\n", ans);
    }

    return 0;
}