luogu P1725 琪露诺

洛谷P1725琪露诺题解

最新推荐文章于 2025-03-15 16:01:25 发布

转载最新推荐文章于 2025-03-15 16:01:25 发布 · 49 阅读

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原文链接：http://www.cnblogs.com/ZlycerQan/p/6747762.html

本文介绍了一种解决洛谷P1725琪露诺问题的方法，采用动态规划与线段树相结合的方式。通过线段树维护最大值，实现了高效的区间查询与更新操作。

二次联通门 : luogu P1725 琪露诺

/*
    luogu P1725 琪露诺 

    DP + 线段树 
    
    用线段树维护dp[i - R] ~ dp[i - L]的最大值
    
    然后 转移方程是    
        dp[i] = max (dp[i - R], dp[i - R + 1], .... dp[i - L - 1], dp[i - L]) + number[i]
    具体边界细节自己乱搞一下就好 
*/
#include <cstdio>

#define Max 200009
#define INF 1e7

inline int max (int a, int b)
{
    return a > b ? a : b;
}

void read (int &now)
{
    now = 0;
    bool flag = false;
    register char word = getchar ();
    while (word > '9' || word < '0')
    {
        if (word == '-')
            flag = true;
        word = getchar ();
    }
    while (word >= '0' && word <= '9')
    {
        now = now * 10 + word - '0';
        word = getchar ();
    }
    if (flag)
        now = -now;
}

int N, L, R;

int number[Max];

struct Segment 
{
    struct Segment_Tree
    {
        int l;
        int r;
        int Maxn;
        int Mid;
    };
    
    Segment_Tree tree[Max << 3];
    
    void Build (int l, int r, int now)
    {
        tree[now].l = l;
        tree[now].r = r;
        if (l == r)
            return ;
        tree[now].Mid = (l + r) >> 1;
        Build (l, tree[now].Mid, now << 1);
        Build (tree[now].Mid + 1, r, now << 1 | 1);
    }
    
    int Query_Maxn (int l, int r, int now)
    {
        if (tree[now].l == l && tree[now].r == r)
            return tree[now].Maxn;
        if (r <= tree[now].Mid)
            return Query_Maxn (l, r, now << 1);
        else if (l > tree[now].Mid)
            return Query_Maxn (l, r, now << 1 | 1);
        else
            return max (Query_Maxn (l, tree[now].Mid, now << 1), Query_Maxn (tree[now].Mid + 1, r, now << 1 | 1));
    }
    
    void Change_Single (int pos, int now, int number)
    {
        if (tree[now].l == tree[now].r)
        {
            tree[now].Maxn = number;
            return ;
        }
        if (pos <= tree[now].Mid)
            Change_Single (pos, now << 1, number);
        else if (pos > tree[now].Mid)
            Change_Single (pos, now << 1 | 1, number);
        tree[now].Maxn = max (tree[now << 1].Maxn, tree[now << 1 | 1].Maxn);
    }
};

Segment Tree;

int main (int argc, char *argv[])
{
    read (N);
    read (L);
    read (R);
    Tree.Build (0, N, 1);
    for (int i = 0; i <= N; i++)
    {
        read (number[i]);
        if (i <= L)
            Tree.Change_Single (i, 1, number[i]); 
    }
    for (int i = L + 1; i <= N; i++)
        if (i - L >= L)
            Tree.Change_Single (i, 1, Tree.Query_Maxn(max(L, i - R), i - L, 1) + number[i]);
        else 
            Tree.Change_Single (i, 1, number[i]);  
    printf ("%d", Tree.Query_Maxn (N - R, N, 1));
    return 0;
}

转载于:https://www.cnblogs.com/ZlycerQan/p/6747762.html