CF-31E-TV Game（dp）_tv31e-优快云博客

本文探讨了在特定游戏规则下，两名玩家如何通过最优的策略分配游戏进程中的数字，以最大化各自的总奖励。利用多阶段决策和动态规划方法解决此问题，详细解释了从后向前整数分配的思路及实现过程。

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E - TV Game

Crawling in process... Crawling failed Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

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Description

There is a new TV game on BerTV. In this game two players get a number A consisting of 2n digits. Before each turn players determine who will make the next move. Each player should make exactly n moves. On it's turn i-th player takes the leftmost digit of A and appends it to his or her number S_i. After that this leftmost digit is erased from A. Initially the numbers of both players (S₁ and S₂) are «empty». Leading zeroes in numbers A, S₁, S₂ are allowed. In the end of the game the first player gets S₁ dollars, and the second gets S₂ dollars.

One day Homer and Marge came to play the game. They managed to know the number A beforehand. They want to find such sequence of their moves that both of them makes exactly n moves and which maximizes their total prize. Help them.

Input

The first line contains integer n (1 ≤ n ≤ 18). The second line contains integer A consisting of exactly 2n digits. This number can have leading zeroes.

Output

Output the line of 2n characters «H» and «M» — the sequence of moves of Homer and Marge, which gives them maximum possible total prize. Each player must make exactly n moves. If there are several solutions, output any of them.

Sample Input

Input

2
1234

Output

HHMM

Input

2
9911

Output

HMHM

思路：每一位不是给W就是给H，很明显是多阶段决策。

有点难想的是从后往前整，先整小的为逐次分配出最优结果。

f[x][y]表示给x位给W，y位给H所得到结果的最优值。

f[x][y]=max(f[x-1][y]+该位给W所增加的值，f[x][y-1]+该位给H所增加的值)

最后把路径输出即可。

失误：DP做的多了，感觉有的有点畏惧感，有点思路，却挺怕不对而花很多时间来整这思路。╮(╯▽╰)╭

这题在大牛眼中应该是水DP吧，不知道我这只菜鸟啥时候也能成大牛啊。

#include<iostream>
#include<cstring>
using namespace std;
const int mm=100;
long long f[mm][mm],w[mm];
int main()
{
  int m;char s[mm];
  while(cin>>m>>s)
  { w[1]=1;
    for(int i=2;i<=18;i++)
      w[i]=w[i-1]*10;
    memset(f,0,sizeof(f));
    for(int i=0;i<=m;i++)
      for(int j=0;j<=m;j++)
    { long long z=s[m+m-i-j]-'0';
      if(i)f[i][j]=f[i-1][j]+w[i]*z;
      if(j)f[i][j]=f[i][j]>f[i][j-1]+w[j]*z?f[i][j]:f[i][j-1]+w[j]*z;
    }
    for(int i=m,j=m;i>0||j>0;)
    { long long z=s[m+m-i-j]-'0';
      if(i&&f[i][j]-f[i-1][j]==z*w[i])
        {--i;cout<<"H";
        }
      else {--j;cout<<"M";}
    }
    cout<<"\n";
  }
}