HDU 5228 ZCC loves straight flush( BestCoder Round #41)

题目链接：

pid=5228">ZCC loves straight flush

pid=5228">题面：

pid=5228">

ZCC loves straight flush

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

Total Submission(s): 827    Accepted Submission(s): 340

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

0
1
2

 

Source

BestCoder Round #41

解题：

注意顺子仅仅能连到A。

代码：

#include <iostream>
#include <cmath>
#include <cstring>
using namespace std;
bool status[4][15];
int main()
{
    int t,n,root;
    cin>>t;
    char c;
    int tmp,cnt,maxx;
    while(t--)
    {
       memset(status,0,sizeof(status));
       for(int i=0;i<5;i++)
       {
            cin>>c>>tmp;
            status[c-'A'][tmp]=1;
             if(tmp==1)status[c-'A'][14]=1;
       }
       maxx=0;
       for(int i=0;i<4;i++)
       {
            for(int j=1;j<=10;j++)
            {
                cnt=0;
            if(status[i][j])cnt++;
              if(status[i][j+1])cnt++;
              if(status[i][j+2])cnt++;
              if(status[i][j+3])cnt++;
              if(status[i][j+4])cnt++;
              if(cnt>maxx)
                maxx=cnt;
         }
       }
       cout<<5-maxx<<endl;
    }
    return 0;
}