HDU 4628 Pieces (状压DP)

Pieces

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 311    Accepted Submission(s): 172

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

 

 

Source

2013 Multi-University Training Contest 3

 

 

Recommend

zhuyuanchen520

 

 

 

题意：

给定一个串，每次可以删除一个回文子串，问把全串删干净的最少次数。

思路：

用状态压缩把所有的状态枚举出来，用数组记录所表示状态所有字符被删除干净的最少次数。

状态：dp[x]表示在状态x下把所有字符删除的最少次数。

状态压缩：dp[x] = min{dp[x], dp[k]}(k是x的子集)

这样的dp的边界不再是dp[0]或dp[n]了。而是对每个状态都要设定原始值，如果某个状态是完全回文的，则dp[x] = 1，否则，dp[x] 就是一个最大值（目前子串的长度）

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>

using namespace std;

const int N=16;
const int INF=0x3f3f3f3f;

char str[30];
int len,dp[1<<16];
vector<int> vt;

int solve(int k){
    vt.clear();
    for(int i=0;i<=len;i++)
        if(k&(1<<i))
            vt.push_back(i);
    int sz=vt.size()-1;
    for(int i=0;i<sz/2;i++)
        if(str[vt[i]]!=str[vt[sz-i]])
            return vt.size();
    return 1;
}

int main(){

    //freopen("input.txt","r",stdin);

    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%s",str);
        len=strlen(str);
        for(int i=1;i<=(1<<len)-1;i++){
            dp[i]=solve(i);
            for(int j=i;j;j=(j-1)&i)
                dp[i]=min(dp[i],dp[j]+dp[j^i]);
        }
        printf("%d\n",dp[(1<<len)-1]);
    }
    return 0;
}