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  • Codeforces Round #605 (Div. 3) D. Remove One Element(DP)

    链接:

    https://codeforces.com/contest/1272/problem/D

    题意:

    You are given an array a consisting of n integers.

    You can remove at most one element from this array. Thus, the final length of the array is n−1 or n.

    Your task is to calculate the maximum possible length of the strictly increasing contiguous subarray of the remaining array.

    Recall that the contiguous subarray a with indices from l to r is a[l…r]=al,al+1,…,ar. The subarray a[l…r] is called strictly increasing if al<al+1<⋯<ar.

    思路:

    正着反着算一边,能往两边延长的最大长度,对每个点特判。

    代码:

    #include<bits/stdc++.h>
    using namespace std;
    const int MAXN = 2e5+10;
    
    int Dp[MAXN], a[MAXN], Dpr[MAXN];
    int n;
    
    int main()
    {
        cin >> n;
        for (int i = 1;i <= n;++i)
            cin >> a[i];
        int ans = 1;
        Dp[0] = Dpr[n+1] = 0;
        Dp[1] = Dpr[n] = 1;
        for (int i = 2;i <= n;++i)
        {
            if (a[i] > a[i-1])
                Dp[i] = Dp[i-1]+1;
            else
                Dp[i] = 1;
        }
        for (int i = n-1;i >= 1;--i)
        {
            if (a[i] < a[i+1])
                Dpr[i] = Dpr[i+1]+1;
            else
                Dpr[i] = 1;
        }
        for (int i = 2;i <= n;++i)
        {
            ans = max(ans, Dp[i]);
            if (a[i] > a[i-2])
                ans = max(ans, Dp[i-2]+Dpr[i]);
        }
        cout << ans << endl;
    
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/YDDDD/p/12046674.html
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