hdu 5288 ZCC loves straight flush

传送门

ZCC loves straight flush

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1038    Accepted Submission(s): 429

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

 

Recommend

hujie   |   We have carefully selected several similar problems for you:  5639 5638 5637 5636 5635 

 

 

思路：

一开始没注意到suit 只能是啊a,b,c,d, 也以为卡片的顺序不能变，，又把问题想复杂了。

其实，只要 哈希花色和数字，然后暴力枚举

 

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <stack>
 #include <cctype>
 #include <vector>
 #include <cmath>
 #include <map>
 #include <queue>

 #define ll long long

 using namespace std;

 int T;
 int mi;
 char s[];
 int t[][];

 void solve()
 {
     int i,j,k;
     int p;
     for(i = ;i < ;i++){
         for(j = ;j <= ;j++ ){
             p = ;
             for(k = ;k <= ;k++){
                 if(t[i][ (j+k-)%+ ]){
                     p++;
                 }
             }
             mi = min(mi, - p);
         }
     }
 }

 int main()
 {
     //freopen("in.txt","r",stdin);
     scanf("%d",&T);
     for(int ccnt=;ccnt<=T;ccnt++){
     //while(scanf("%d%s%s",&n,a,b)!=EOF){
         mi = ;
         int i;
         int v;
         memset(t,,sizeof(t));
         for(i = ;i <= ;i++){
             scanf("%s",s);
             v = s[] - '';
             if(strlen(s) == ){
                 v*=;
                 v+=s[] - '';
             }
             t[ s[] - 'A' ][ v ] = ;
         }
         solve();
         printf("%d\n",mi);
     }
     return ;
 }