[动态规划]状态压缩DP小结

 1.小技巧

枚举集合S的子集:for(int i = S; i > 0; i=(i-1)&S)

枚举包含S的集合:for(int i = S; i < (1<<n); i ＝ (i+1)|S)

2.训练题目

1.HDU 4628 Pieces http://acm.hdu.edu.cn/showproblem.php?pid=4628

题意: 给定一个(len<=16)的串,每次可删除一个回文子序列,问最少需要几次删完整个串

思路:简单状态压缩   dp[s]=min(dp[s],dp[s^sub]+1)  (sub为s的回文子序列)

# include<map>
# include<set>
# include<cmath>
# include<queue>
# include<stack>
# include<vector>
# include<string>
# include<cstdio>
# include<cstring>
# include<iostream>
# include<algorithm>
# include<functional>
using namespace std;

typedef pair<int,int> PII;
# define lson l,m,rt<<
# define rson m+,r,rt<<|
# define MOD
# define LL long long
# define pb push_back
# define F first
# define S second
# define N
# define M
char s[N];
int len;
int flag[M],dp[M];

int check(int mask)
{
    int i,n;
    string sub="";
    for(i=;i<len;i++)
        if(mask&(<<i))
            sub+=s[i];
    n=sub.length();
    for(i=;i<n/;i++)
        if(sub[i]!=sub[n--i])
            return ;
    return ;
}

int dfs(int mask)
{
    int i,ans=<<;
    if(!mask) return ;
    if(dp[mask]!=-) return dp[mask];
    for(i=mask;i>;i=(i-)&mask)   //枚举mask的所有子集
        if(flag[i])
            ans=min(ans,dfs(mask^i));
    return dp[mask]=ans+;
}

int main()
{
    //freopen("in.txt","r",stdin);
    int i,T,ans;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",s);
        len=strlen(s);
        memset(dp,-,sizeof(dp));
        memset(flag,,sizeof(flag));   //是否为回文
        for(i=;i<(<<len);i++)
            flag[i]=check(i);
        ans=dfs((<<len)-);
        printf("%d\n",ans);
    }
    return ;
}

HDU 4628