「USACO16OPEN」「LuoguP3146」248（区间dp

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 NNN positive integers (2≤N≤2482 leq Nleq 2482≤N≤248), each in the range 1…401 ldots 401…40. In one move, Bessie cantake two adjacent numbers with equal values and replace them a singlenumber of value one greater (e.g., she might replace two adjacent 7swith an 8). The goal is to maximize the value of the largest numberpresent in the sequence at the end of the game. Please help Bessiescore as highly as possible!

给定一个1*n的地图，在里面玩2048，每次可以合并相邻两个（数值范围1-40），问最大能合出多少。注意合并后的数值并非加倍而是+1，例如2与2合并后的数值为3。

输入输出格式

输入格式：

The first line of input contains NNN, and the next NNN lines give the sequence

of NNN numbers at the start of the game.

输出格式：

Please output the largest integer Bessie can generate.

输入输出样例

输入样例#1： 复制

4
1
1
1
2

输出样例#1： 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.

---

题解

一个还算朴素的区间dp。

设f[i][j]为从i到j全部合起来的最大值，枚举断点k，转移方程：

if(f[i][k]==f[k+1][j])
  f[i][j]=f[i][k]+1;

然后注意一下for循环的嵌套顺序就星了。

/*
    qwerta
    P3146 [USACO16OPEN]248
    Accepted
    100
    代码 C++，0.5KB
    提交时间 2018-09-18 16:19:46
    耗时/内存
    115ms, 1040KB
*/
#include<cmath>
#include<cstdio>
#include<iostream>
using namespace std;
int f[257][257];
int main()
{
    //freopen("a.in","r",stdin);
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;++i)
    scanf("%d",&f[i][i]);
    for(int i=n-1;i;--i)
    for(int j=i+1;j<=n;++j)
    {
        for(int k=i;k<j;++k)
          if(f[i][k]==f[k+1][j])
            f[i][j]=f[i][k]+1;
    }
    int ans=0;
    for(int i=1;i<=n;++i)
    for(int j=i;j<=n;++j)
    ans=max(ans,f[i][j]);
    cout<<ans;
    return 0;
}