【bzoj4580】[Usaco2016 Open]248 区间dp

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 g

ame she is playing. The game starts with a sequence of NN positive integers (2≤N≤248), 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

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

Please output the largest integer Bessie can generate.

样例输入

4
1
1
1
2

样例输出

3

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题目大意

给你n个数的数列，每次可以合并两个相邻且相等的数，变成大小比原数大1的新数。问可能出现的最大的数是多少。

题解

区间dp

首先必须明确一点，如果一段区间能够按照题目要求合并成一个数，那么这个数一定是确定的。

那么可以设f[i][j]为i到j这段区间合并成的数，如果无法合并则为0。

那么就有f[i][j]=f[i][k]+1(f[i][k]==f[k+1][j]&&f[i][k]!=0)。

答案就是max{f[i][j]}。

注意需要先枚举长度再枚举起始点。

#include <cstdio>
#include <algorithm>
using namespace std;
int f[250][250];
int main()
{
	int n , i , j , k , ans = 0;
	scanf("%d" , &n);
	for(i = 1 ; i <= n ; i ++ )
		scanf("%d" , &f[i][i]);
	for(i = 2 ; i <= n ; i ++ )
		for(j = 1 ; j <= n - i + 1 ; j ++ )
			for(k = j ; k <= i + j - 2 ; k ++ )
				if(f[j][k] && f[k + 1][i + j - 1] && f[j][k] == f[k + 1][i + j - 1])
					f[j][i + j - 1] = f[j][k] + 1;
	for(i = 1 ; i <= n ; i ++ )
		for(j = i ; j <= n ; j ++ )
			ans = max(ans , f[i][j]);
	printf("%d\n" , ans);
	return 0;
}

 

转载于:https://www.cnblogs.com/GXZlegend/p/6434903.html