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1046 [USACO 2016 Ope P]262144 区间DP 不会~~~~~~~~~~~~~

开发技术 开发技术 2022-07-17 次浏览

https://blog.csdn.net/qq_45516669/article/details/108209561?spm=1001.2101.3001.6661.1&utm_medium=distribute.pc_relevant_t0.none-task-blog-2%7Edefault%7EBlogCommendFromBaidu%7Edefault-1-108209561-blog-121525948.pc_relevant_sortByStrongTime&depth_1-utm_source=distribute.pc_relevant_t0.none-task-blog-2%7Edefault%7EBlogCommendFromBaidu%7Edefault-1-108209561-blog-121525948.pc_relevant_sortByStrongTime&utm_relevant_index=1

 

 

链接:https://ac.nowcoder.com/acm/contest/24213/1046
来源:牛客网

题目描述

Bessie likes downloading games to play on her cell phone, even though she does find the small touch screen rather cumbersome to use with her large hooves.
She is particularly intrigued by the current game she is playing. The game starts with a sequence of N positive integers (2≤N≤262,144), each in the range 1…40. In one move, Bessie can take two adjacent numbers with equal values and replace them a single number of value one greater (e.g., she might replace two adjacent 7s with an 8). The goal is to maximize the value of the largest number present in the sequence at the end of the game. Please help Bessie score as highly as possible!

输入描述:

The first line of input contains N, and the next N lines give the sequence of N numbers at the start of the game.

输出描述:

Please output the largest integer Bessie can generate.
示例1

输入

复制
4
1
1
1
2

输出

复制
3

说明

In this example shown here, Bessie first merges the second and third 1s to obtain the sequence 1 2 2, and then she merges the 2s into a 3. Note that it is not optimal to join the first two 1s.



分析

题意:一段数组内,任意两个相邻的相等的数都可以合并成一个更大的数,问最终合并的最大值是多少

普通的区间DP是比较好想的,但是要n^3 ,就不会了。

设dp[i][j] 表示 i 到j 区间内能合并的最大值,转换方程:dp[i][k] == dp[k][j] 的时候,dp[i][j] = max(dp[i][j],dp[i][k] + dp[k][j] + 1)  

一道很难的dp题,涉及到指针的转换,值的转换。是一道区间DP,具体看上面题解,我实力不够,或者有想法比较透彻的评论区教教我,厚礼蟹。

 

普通code:

#include<bits/stdc++.h>
using namespace std;
int n;
int f[250];
int dp[250][250];
inline int read()
{
    int Num=0,f=1; char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-') f=-1; ch=getchar();}
    while(ch>='0'&&ch<='9') {Num=(Num<<1)+(Num<<3)+ch-'0'; ch=getchar();}
    return Num*f;
}
int main()
{
    n=read();
    for(int i=1;i<=n;i++)
    {
        f[i]=read();
        dp[i][i]=f[i];
    }
    int maxx=0;
    for(int len=1;len<=n;len++)
        for(int l=1;l<=n-len+1;l++)
        {
            int r=l+len-1;
            for(int k=l;k<r;k++)
                if(dp[l][k]==dp[k+1][r])
                {
                    dp[l][r]=max(dp[l][r],dp[l][k]+1);
                    maxx=max(maxx,dp[l][r]);
                }
        }
    printf("%d ",maxx);
    return 0;
}

 

题解code:

#include<bits/stdc++.h>
using namespace std;
int n,ans;
int dp[60][270007];
inline int read()
{
    int Num=0,f=1; char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-') f=-1; ch=getchar();}
    while(ch>='0'&&ch<='9') {Num=(Num<<1)+(Num<<3)+ch-'0'; ch=getchar();}
    return Num*f;
}
int main()
{
    n=read(); 
    for(int i=1;i<=n;i++) dp[read()][i]=i+1;
    for(int i=2;i<=58;i++)
    for(int j=1;j<=n;j++)
    {
        if(!dp[i][j]) dp[i][j]=dp[i-1][dp[i-1][j]];
        if(dp[i][j]) ans=i;
    }
    cout<<ans;
    return 0;
}

 

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