Card Hand Sorting

   时间限制: 1 Sec  内存限制: 64 MB
提交: 60  解决: 22
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When dealt cards in the card game Plump it is a good idea to start by sorting the cards in hand by suit and rank. The different suits should be grouped and the ranks should be sorted within each suit. But the order of the suits does not matter and within each suit, the cards may be sorted in either ascending or descending order on rank. It is allowed for some suits to be sorted in ascending order and others in descending order.
Sorting is done by moving one card at a time from its current position to a new position in the hand, at the start, end, or in between two adjacent cards. What is the smallest number of moves required to sort a given hand of cards?

输入

The ﬁrst line of input contains an integer n (1 ≤ n ≤ 52), the number of cards in the hand. The second line contains n pairwise distinct space-separated cards, each represented by two characters. The ﬁrst character of a card represents the rank and is either a digit from 2 to 9 or
one of the letters T , J , Q , K , and A representing Ten, Jack, Queen, King and Ace, respectively, given here in increasing order. The second character of a card is from the set { s , h , d , c } representing the suits spades ♠, hearts ♥, diamonds ♦, and clubs ♣.

输出

Output the minimum number of card moves required to sort the hand as described above.

样例输入

7
9d As 2s Qd 2c Jd 8h

样例输出

题意
给出一个扑克牌的序列求排成一个“有序”序列最少的插入次数
有序是这样定义的同一个花色的必须放在一起

同一花色中的牌必须是升序或者是降序

然后 A 是最大的（A 是 14 不是 1）坑啊

思路

我们可以枚举一下
一共有四种花色的就是 4!
每种升序有升序和降序就是 2 ^ 4
4 ! * 2 ^ 4 = 384 可以接受
然后次数怎么求呢就是 n - LCS

因为这不是交换来变得有序如果是交换来变得有序就是求逆序数
插入的话就将这个序列求出其最长上升子列这个子列不动其他元素依次插入就可以

然后讲讲怎么枚举

首先枚举花色就是
我们将四种花色分别与 0 1 2 3 对应起来
然后用一个数组 color[4]= {0, 1, 2, 3} 求这个花色的全排列这就定义了花色的先后次序

color[i] 表示数字i 对应的花色的优先级是多少

然后枚举升序降序

我们可以用 0-15 这16 个数字的二进制来表示升序还是降序

我们可以重新给它定义一个 tv
降序就是将原值取负数然后按升序排序实际上就是原值的降序排序

比如 0 这个数字对应四位的二进制数字是 0000

所以这个时候四种花色都是升序的我们可以用花色的优先级来表示对应哪位数字

比如此时 s 花色对应的优先级是 0 那么就是将 0 右移 0 位再 & 1 就可以判断它对应位是 1 还是 0

#include<bits/stdc++.h>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
struct node
{
    int id;
    int colr;
    int num;
    int tv;
}p[150];
int color[4]={0,1,2,3};
int n;
int cmp(node a,node b)
{
    if(a.colr==b.colr)
        return a.tv<b.tv;

    return color[a.colr]<color[b.colr];


}
char s[150];
int LCS()
{
    int dp[60];
    memset(dp,0,sizeof(dp));
    dp[0] = 1;
    int ans = 1;
    for (int i = 1; i < n; i++)
    {
        dp[i] = 1;
        for (int j = 0; j < i; j++)
        {
            if (p[i].id > p[j].id)
            {
                dp[i] = max(dp[i], dp[j] + 1);
            }
        }
        ans = max(dp[i], ans);
    }
    return ans;
}

int main()
{

    int ans=inf;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%s",s);
        if(s[0]=='T')p[i].num=10;
        else if(s[0]=='J')p[i].num=11;
        else if(s[0]=='Q')p[i].num=12;
        else if(s[0]=='K')p[i].num=13;
        else if(s[0]=='A')p[i].num=14;
        else p[i].num=s[0]-'0';

        if(s[1]=='s')p[i].colr=0;
        else if(s[1]=='h')p[i].colr=1;
        else if(s[1]=='d')p[i].colr=2;
        else if(s[1]=='c')p[i].colr=3;
        p[i].id=i;
    }
    do
    {
        for(int i=0;i<16;i++)
        {
            for(int j=0;j<n;j++)
            {
                p[j].tv=p[j].num*(((i>>color[p[j].colr])&1) ? 1 : -1);

            }
            sort(p,p+n,cmp);
            int minn=n-LCS();
            ans=min(minn,ans);
        }
    }while(next_permutation(color,color+4));
    printf("%d\n",ans);
    return 0;
}

Card Hand Sorting(枚举排列+最长公共子序列)

Card Hand Sorting

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