Problem Solving►UVa

UVa 10534 - Wavio Sequence

Contents

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

Problem

中文網址

Solution

LIS 方法可參考 http://www.csie.ntnu.edu.tw/~u91029/LongestIncreasingSubsequence.html#3

做 LIS 和 LDS，比較每個數字在 LIS 串和 LDS 串中的位置，挑選位置相同且最長的。

    1 2 3 4 5 4 3 2 1 10

LIS 1 2 3 4 5 4 3 2 1 6
LDS 1 2 3 4 5 4 3 2 1 1

其中最長的會是 min(LIS[4],LDS[4]) = 5，而其長度就是 5 * 2 - 1 = 9

這邊 LDS 的做法是採取從後面往前做 LIS。皆為 $O(nlogn)$ 。

Code

UVa 10534
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#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;

int main()
{
    int n, num[10000], lis[10000], lds[10000];
    vector<int> v;
    while (scanf("%d", &n) != EOF)
    {
        int i;
        for (i = 0; i < n; i++)
            scanf("%d", &num[i]);

        v.clear();
        int size = 1;
        v.push_back(num[0]);
        lis[0] = 1;
        for (i = 1; i < n; i++)
        {
            if (num[i] > v.back())
            {
                v.push_back(num[i]);
                lis[i] = ++size;
            }
            else
            {
                int temp = lower_bound(v.begin(), v.end(), num[i]) - v.begin();
                v[temp] = num[i];
                lis[i] = temp + 1;
            }
        }

        v.clear();
        size = 1;
        v.push_back(num[n - 1]);
        lds[n - 1] = 1;
        for (i = n - 2; i >= 0; i--)
        {
            if (num[i] > v.back())
            {
                v.push_back(num[i]);
                lds[i] = ++size;
            }
            else
            {
                int temp = lower_bound(v.begin(), v.end(), num[i]) - v.begin();
                v[temp] = num[i];
                lds[i] = temp + 1;
            }
        }

        /*for (i = 0; i < n; i++)
            printf("%d ", lis[i]);
        putchar('\n');
        for (i = 0; i < n; i++)
            printf("%d ", lds[i]);*/

        int max = 0;
        for (i = 0; i < n; i++)
        {
            int temp = min(lis[i], lds[i]);
            if (max < temp)
                max = temp;
        }

        printf("%d\n", max * 2 - 1);
    }

    return 0;
}