P3147 [USACO16OPEN]262144 P

[USACO16OPEN]262144 P

题目描述

Bessie likes downloading games to play on her cell phone, even though she doesfind 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 (\(2 \leq N\leq 262,144\)), each in the range \(1 \ldots 40\). In one move, Bessiecan take two adjacent numbers with equal values and replace them asingle number of value one greater (e.g., she might replace twoadjacent 7s with an 8). The goal is to maximize the value of thelargest number present in the sequence at the end of the game. Pleasehelp Bessie score as highly as possible!

Bessie喜欢在手机上下游戏玩（……），然而她蹄子太大，很难在小小的手机屏幕上面操作。

她被她最近玩的一款游戏迷住了，游戏一开始有n个正整数，(2<=n<=262144)，范围在1-40。在一步中，贝西可以选相邻的两个相同的数，然后合并成一个比原来的大一的数（例如两个7合并成一个8)，目标是使得最大的数最大，请帮助Bessie来求最大值。

输入格式

The first line of input contains \(N\), and the next \(N\) lines give the sequence

of \(N\) numbers at the start of the game.

输出格式

Please output the largest integer Bessie can generate.

样例 #1

样例输入 #1

4
1
1
1
2

样例输出 #1

3

提示

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.

思路

我们观察262144
在计算机上按出来

居然是2的18次方
我们思考能不能把这个log用上
我们常规的状态设置都是左右端点
这次我们换一下因为值域正好很小40+18也很小
我们设置f[i][j]表示左端点为j 能合成i 的右端点值
我们思考状态转移：
显然我们合成i 是从 i-1转移
f[i][j]=max{f[i-1][f[i-1][j] +1 ]}
注意这里要加1 因为我们f[i-1][j]表示的右端点 而我们放进去这个状态表示是左端点的
这样我们的状态转移就出来了
if(dp[i-1][j]&&dp[i-1][dp[i-1][j]+1])dp[i][j]=max(dp[i][j],dp[i-1][dp[i-1][j]+1]);
当然我们要是没有初始化的就是不合法状态 直接不让他转移即可

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5+10;
const int M = 998244353;
const int mod = 1e9+7;
#define int long long
#define endl '\n'
#define all(x) (x).begin(),(x).end()
#define YES cout<<"YES"<<endl;
#define NO cout<<"NO"<<endl;
#define _ 0
#define pi acos(-1)
#define INF 0x3f3f3f3f3f3f3f3f
#define fast ios::sync_with_stdio(false);cin.tie(nullptr);
int dp[59][300010];
void solve() {
    int n;cin>>n;
    vector<int>a(n+1);
    for(int i=1;i<=n;i++){
        int x;cin>>x;
        dp[x][i]=i;
    }
    int ans=0;
    for(int i=1;i<=58;i++)
        for(int j=1;j<=n;j++){
            if(dp[i-1][j]&&dp[i-1][dp[i-1][j]+1])dp[i][j]=max(dp[i][j],dp[i-1][dp[i-1][j]+1]);
            if(dp[i][j])ans=max(ans,i);
        }
    cout<<ans<<endl;
}
signed main(){
    fast
    int T;T=1;
    while(T--) {
        solve();
    }
    return ~~(0^_^0);
}