[USACO16OPEN]262144

题目链接：点这里

Solution:

设\(f[i][j]\)表示以j为左端点，合并后最大值为i的序列的右端点的位置

那么我们可以得到状态转移方程\(f[i][j]=f[i-1][f[i-1][j]+1]\)

因为\(N\le262144\)，所以i最大可能为58，那么我们枚举即可

Code：

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=262146;
int n,ans,f[59][N];
int read(){int x=0,f=1;char ch=getchar();while(!isdigit(ch)){if(ch=='-')f=-f;ch=getchar();}while(isdigit(ch)){x=x*10+ch-48;ch=getchar();}return x*f;
}
int main(){n=read();for(int i=1;i<=n;i++) f[read()][i]=i;for(int i=1;i<=58;i++) for(int j=1;j<=n;j++){!f[i][j]&&f[i-1][j]&&(f[i][j]=f[i-1][f[i-1][j]+1]);f[i][j]?ans=i:0;}printf("%d\n",ans);return 0;
}

转载于:https://www.cnblogs.com/NLDQY/p/10919990.html

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[USACO16OPEN]262144

Solution:

Code：

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