hdu5228 ZCC loves straight flush(模拟)

最新推荐文章于 2016-09-24 16:07:03 发布

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最新推荐文章于 2016-09-24 16:07:03 发布

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文章标签： hdu5228 模拟

本文链接：https://blog.youkuaiyun.com/yuyanggo/article/details/45878057

本文探讨了在德州扑克游戏中通过技术手段调整牌面，以达到同花顺的最佳组合，详细介绍了计算所需更改的最小牌张数量的方法，并强调了同花顺中牌序的重要性。

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ZCC loves straight flush

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)

Total Submission(s): 611 Accepted Submission(s): 266

Problem Description

After losing all his chips when playing Texas Hold'em with Fsygd on the way to ZJOI2015, ZCC has just learned a black technology. Now ZCC is able to change all cards as he wants during the game. ZCC wants to get a Straight Flush by changing as few cards as possible.

We call a five-card hand a Straight Flush when all five cards are consecutive and of the same suit. You are given a five-card hand. Please tell ZCC how many cards must be changed so as to get a Straight Flush.

Cards are represented by a letter('A', 'B', 'C', 'D') which denotes the suit and a number('1', '2',

⋯ , '13') which denotes the rank.

Note that number '1' represents ace which is the largest actually. "1 2 3 4 5" and "10 11 12 13 1" are both considered to be consecutive while "11 12 13 1 2" is not.

Input

First line contains a single integer

T(T=1000) which denotes the number of test cases.
For each test case, there are five short strings which denote the cards in a single line. It's guaranteed that all five cards are different.

Output

For each test case, output a single line which is the answer.

Sample Input

  
   3
A1 A2 A3 A4 A5
A1 A2 A3 A4 C5
A9 A10 C11 C12 C13

Sample Output

Source

BestCoder Round #41

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解析：模拟。直接暴力枚举所有的胜利情况，然后一一匹配，得到匹配牌数x，需要改变的牌就为5-x。

需要注意的是，5招牌并不一定是有序的，也就是说：A1 A2 A5 A4 A3 也是同花顺。一开始没注意到这个，调试一下午都没找出来。

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 10
using namespace std;

int a[maxn],b[maxn];
char s[maxn];

void work()
{
  int i,j,k,p,q,sum,ans,x,y;
  for(i=1;i<=5;i++)
    {
      scanf("%s",s);
      a[i]=s[0]-'A';
      b[i]=s[1]-'0';
      k=strlen(s);
      for(j=2;j<k;j++)b[i]=b[i]*10+s[j]-'0';
    }
  
  ans=5;
  for(i=0;i<4;i++)
    for(j=1;j<=10;j++)
      {
        sum=0,x=i,y=j;
        for(k=0;k<5;k++)
          {
            if(y==14)y=1;
            for(p=1;p<=5;p++)
              if(a[p]==x && b[p]==y)sum++;
            y++;  
          }
        ans=min(ans,5-sum);  
      }
  printf("%d\n",ans);	  
}

int main()
{
  freopen("1.in","r",stdin);
  int t,i,j,k;
  while(scanf("%d",&t)!=EOF)
    for(i=1;i<=t;i++)work();
  return 0; 
}