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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): 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
 
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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 ;
}

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