Luogu P3609 [USACO17JAN]Hoof, Paper, Scissor蹄子剪刀... (dp)

题目背景

欢迎提供翻译，请直接在讨论区发帖，感谢你的贡献。

题目描述

You have probably heard of the game "Rock, Paper, Scissors". The cows like to play a similar game they call "Hoof, Paper, Scissors".

The rules of "Hoof, Paper, Scissors" are simple. Two cows play against each-other. They both count to three and then each simultaneously makes a gesture that represents either a hoof, a piece of paper, or a pair of scissors. Hoof beats scissors (since a hoof can smash a pair of scissors), scissors beats paper (since scissors can cut paper), and paper beats hoof (since the hoof can get a papercut). For example, if the first cow makes a "hoof" gesture and the second a "paper" gesture, then the second cow wins. Of course, it is also possible to tie, if both cows make the same gesture.

Farmer John wants to play against his prize cow, Bessie, at $N$ games of "Hoof, Paper, Scissors" ( $1 \leq N \leq 100,000$ ). Bessie, being an expert at the game, can predict each of FJ's gestures before he makes it. Unfortunately, Bessie, being a cow, is also very lazy. As a result, she tends to play the same gesture multiple times in a row. In fact, she is only willing to switch gestures at most $K$ times over the entire set of games ( $0 \leq K \leq 20$ ). For example, if $K=2$ , she might play "hoof" for the first few games, then switch to "paper" for a while, then finish the remaining games playing "hoof".

Given the sequence of gestures FJ will be playing, please determine the maximum number of games that Bessie can win.

你可能听说过“石头，剪刀，布”的游戏。FJ的牛喜欢玩一个类似的游戏，它们称之为“蹄子，剪刀，布”（“蹄子”就是“石头”）。

游戏规则很简单：比赛双方同时数到3，然后同时出一个手势，代表“蹄子”“剪刀”或“布”。“蹄子”胜“剪刀”，“剪刀”胜“布”，“布”胜“蹄子”。举个例子，第一头牛出“蹄子”，第二头牛出“布”，则第二头牛胜利。当然，也可以“平局”（如果两头牛手势相同的话）。

FJ想对阵自己获奖的牛，贝西。贝西作为一个专家，能够预测FJ的手势。不幸的是，贝西作为一头牛，也十分的懒惰。事实上，她只愿意变换固定次数的手势来完成游戏。例如，她可能只想变1次，则他可能出“蹄子”几次，剩下的都出“布”。

鉴于贝西预测FJ会出的手势，以及她想变的次数，求出她最多能赢多少场

输入输出格式

输入格式：

The first line of the input file contains $N$ and $K$ .

The remaining $N$ lines contains FJ's gestures, each either H, P, or S.

输出格式：

Print the maximum number of games Bessie can win, given that she can only change gestures at most $K$ times.

输入输出样例

输入样例#1：

5 1
P
P
H
P
S

输出样例#1：

4

说明

感谢 @lzyzz250 的翻译

---

看题目不难想到这是个dp题
然后显然有三个变量来描述一个状态：手势 换了的次数 局数
描述的变量就是当前状态下胜利数最大值
然后从前向后递推就ok
dp[i][j][h]={max⁡{dp[i][j][h],dp[k][j][h−1]}i==k,win[h]!=imax⁡{dp[i][j][h],dp[k][j][h−1]+1}i==k,win[h]==imax⁡{dp[i][j][h],dp[k][j−1][h−1]}i!=k,j>0,win[h]!=imax⁡{dp[i][j][h],dp[k][j−1][h−1]+1}i!=k,j>0,win[h]==idp[i][j][h] = \begin{cases} \max\{dp[i][j][h],dp[k][j][h-1]\}&&i==k,win[h]!=i\\ \max\{dp[i][j][h],dp[k][j][h-1]+1\}&&i==k,win[h]==i\\ \max\{dp[i][j][h],dp[k][j-1][h-1]\}&&i!=k,j>0,win[h]!=i\\ \max\{dp[i][j][h],dp[k][j-1][h-1]+1\}&&i!=k,j>0,win[h]==i\\ \end{cases}dp[i][j][h]=⎩⎪⎪⎪⎨⎪⎪⎪⎧​max{dp[i][j][h],dp[k][j][h−1]}max{dp[i][j][h],dp[k][j][h−1]+1}max{dp[i][j][h],dp[k][j−1][h−1]}max{dp[i][j][h],dp[k][j−1][h−1]+1}​​i==k,win[h]!=ii==k,win[h]==ii!=k,j>0,win[h]!=ii!=k,j>0,win[h]==i​
上面这个东西看起来很麻烦，但是其实没有这么麻烦。。。
代码

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#define For(i,l,r) for(int i=l;i<=r;++i)
#define MAXN 100010
using namespace std;//H=1 P=2 S=3
const int win[4]={0,2,3,1};
int n,k,ges[MAXN],dp[4][21][MAXN],ans;
int main()
{
	char c;
	scanf("%d %d",&n,&k);
	For(i,1,n)
	{
		while((c=getchar())==' '||c=='\n'||c=='\r');
		if(c=='H')
		 ges[i]=1;
		else if(c=='P')
		 ges[i]=2;
		else
		 ges[i]=3;
	}
	For(i,1,n)
	{
		For(j,1,3)//手势 
		{
			For(h,1,3)
			{
				For(l,0,k)
				{
					if(j==h)//不换 
					{
						dp[j][l][i]=max(dp[j][l][i],dp[h][l][i-1]);
					}
					else//换 
					{
						if(l)//够换 
						{
							dp[j][l][i]=max(dp[j][l][i],dp[h][l-1][i-1]);
						}
					}
				}
			}
		}
		For(j,0,k)
		{
			dp[win[ges[i]]][j][i]++;
		}
	}
	For(i,1,3)
	 For(j,0,k)
	  ans=max(ans,dp[i][j][n]);
	printf("%d",ans);
	return 0;
}