hdu 5288 ZCC loves straight flush

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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

BestCoder Round #41

 

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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[10];
int t[5][20];

void solve()
{
    int i,j,k;
    int p;
    for(i = 0;i < 4;i++){
        for(j = 1;j <= 10;j++ ){
            p = 0;
            for(k = 0;k <=4 ;k++){
                if(t[i][ (j+k-1)%13+1 ]){
                    p++;
                }
            }
            mi = min(mi,5 - p);
        }
    }
}

int main()
{
    //freopen("in.txt","r",stdin);
    scanf("%d",&T);
    for(int ccnt=1;ccnt<=T;ccnt++){
    //while(scanf("%d%s%s",&n,a,b)!=EOF){
        mi = 4;
        int i;
        int v;
        memset(t,0,sizeof(t));
        for(i = 1;i <= 5;i++){
            scanf("%s",s);
            v = s[1] - '0';
            if(strlen(s) == 3){
                v*=10;
                v+=s[2] - '0';
            }
            t[ s[0] - 'A' ][ v ] = 1;
        }
        solve();
        printf("%d
",mi);
    }
    return 0;
}