hdu 4628 Pieces（状态压缩＋记忆化搜索）

Pieces

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1811    Accepted Submission(s): 932

Problem Description

You heart broke into pieces.My string broke into pieces.But you will recover one day,and my string will never go back.
Given a string s.We can erase a subsequence of it if this subsequence is palindrome in one step. We should take as few steps as possible to erase the whole sequence.How many steps do we need?
For example, we can erase abcba from axbyczbea and get xyze in one step.

 

Input

The first line contains integer T,denote the number of the test cases. Then T lines follows,each line contains the string s (1<= length of s <= 16).
T<=10.

 

Output

For each test cases,print the answer in a line.

 

Sample Input

2
aa
abb

 

Sample Output

1
2

 

Author

WJMZBMR

 

Source

2013 Multi-University Training Contest 3

题意：给一个字符串，长度<＝16,每次去掉一个回文串。能够中不连续的。问最少用多少次把所给的串都去掉。

解题：先用状态压缩把回文串的状态标记。再用记忆化搜索或dp。

#include<stdio.h>
#include<string.h>
const int inf=20;
int dp[1<<17],len,flag[1<<17];
char str[20];
bool judge(int sta){
    char s[20];
    int k=0 , l , r;
    for(int i=0; (1<<i)<=sta; i++)
        if((1<<i)&sta)
        s[k++]=str[i];
    if(k==0)return 0;
    if(k==1)return 1;
    if(k&1){
        l=k/2-1; r=l+2;
    }
    else{
        l=k/2-1; r=l+1;
    }
    while(r<k&&s[l]==s[r])l--,r++;
    if(r<k)return 0;
    else return 1;
}
void dfs(int sta){
    if(dp[sta]!=inf)
        return ;
    for(int s=sta-1;s>0; s=(s-1)&sta){
        if(!flag[s^sta])continue;
        dfs(s);
        if(dp[sta]>dp[s]+1)
            dp[sta]=dp[s]+1;
    }
    if(judge(sta))
        if(dp[sta]>1)
            dp[sta]=1;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%s",str);
        len=strlen(str);
        for(int i=(1<<len)-1; i>0; i--){
            dp[i]=inf;
            flag[i]=judge(i);
        }
        dp[0]=0;flag[0]=0;

        dfs((1<<len)-1);
        printf("%d
",dp[(1<<len)-1]);
    }
    return 0;
}