Card Hand Sorting
时间限制: 1 Sec 内存限制: 64 MB提交: 66 解决: 23
[提交] [状态] [讨论版] [命题人:外部导入]
题目描述
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 ♣.
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
2
题意
去掉的大小王的扑克牌，要求对给出扑克排序，你每次可以拿出一张牌，插到任意位置，重复此操作直到序列有序。有序定义为：相同花色必须在一起，相同花色必须升序或者降序，不同的花色之间没有要求。
分析
- 总共四种花色，每种花色的牌的摆放顺序只有两种，我们显然可以枚举出最终所有的有序状态：
①按花色的先后顺序有4！共有24种可能。
②按每种花色的排序方式共有2⁴种可能。 - 我们求解初始状态到各个最终状态之间个最少操作数，更新最小值就是结果。
- 问题转换为：给定两个序列，求按上述调整方法由一个序列得到另外一个序列需要的最少操作数。
- 我这里没有用4中提到的方法，而是直接求两个序列的最长公共子序列，然后总序列长度减去子序列的长度就是需要操作的最少次数。
/// author:Kissheart /// #include<stdio.h> #include<algorithm> #include<iostream> #include<string.h> #include<vector> #include<stdlib.h> #include<math.h> #include<queue> #include<deque> #include<ctype.h> #include<map> #include<set> #include<stack> #include<string> #define INF 0x3f3f3f3f #define FAST_IO ios::sync_with_stdio(false) const double PI = acos(-1.0); const double eps = 1e-6; const int MAX=1e5+10; const int mod=1e9+7; typedef long long ll; using namespace std; #define gcd(a,b) __gcd(a,b) inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;} inline ll qpow(ll a,ll b){ll r=1,t=a; while(b){if(b&1)r=(r*t)%mod;b>>=1;t=(t*t)%mod;}return r;} inline ll inv1(ll b){return qpow(b,mod-2);} inline ll exgcd(ll a,ll b,ll &x,ll &y){if(!b){x=1;y=0;return a;}ll r=exgcd(b,a%b,y,x);y-=(a/b)*x;return r;} inline ll read(){ll x=0,f=1;char c=getchar();for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;for(;isdigit(c);c=getchar()) x=x*10+c-'0';return x*f;} //freopen( "in.txt" , "r" , stdin ); //freopen( "data.txt" , "w" , stdout ); int dp[55][55],n; int order[4]={0,1,2,3}; char a[2]; map<char,int>mp; vector<int>v[5],s,t; int main() { mp['s']=0;mp['h']=1;mp['d']=2;mp['c']=3;mp['2']=0;mp['3']=1;mp['4']=2;mp['5']=3;mp['6']=4; mp['7']=5;mp['8']=6;mp['9']=7;mp['T']=8;mp['J']=9;mp['Q']=10;mp['K']=11,mp['A']=12; s.push_back(0); t.push_back(0); scanf("%d",&n); for(int i=1;i<=n;i++) { scanf("%s",a); s.push_back(mp[a[0]]+mp[a[1]]*13); v[mp[a[1]]].push_back(mp[a[0]]+mp[a[1]]*13); } sort(v[0].begin(),v[0].end()); sort(v[1].begin(),v[1].end()); sort(v[2].begin(),v[2].end()); sort(v[3].begin(),v[3].end()); int ans=INF; do { for(int i=0;i<16;i++) { for(int j=0;j<4;j++) { if(!v[order[j]].size()) continue; if((i>>j)&1) { for(int k=0;k<v[order[j]].size();k++) t.push_back(v[order[j]][k]); } else { for(int k=v[order[j]].size()-1;k>=0;k--) t.push_back(v[order[j]][k]); } } memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { if(s[i]==t[j]) dp[i][j]=dp[i-1][j-1]+1; else dp[i][j]=max(dp[i-1][j],dp[i][j-1]); } } ans=min(ans,n-dp[n][n]); t.clear();t.push_back(0); } }while(next_permutation(order,order+4)); printf("%d ",ans); return 0; }