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

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.`

```4
1
1
1
2```

`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

```#include<bits/stdc++.h>
using namespace std;
int n;
int f[250];
int dp[250][250];
{
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()
{
for(int i=1;i<=n;i++)
{
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;
}```

```#include<bits/stdc++.h>
using namespace std;
int n,ans;
int dp[60][270007];
{
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()
{
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;
}```