状态压缩Dp

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本文探讨了如何通过动态规划解决字符串子序列回文问题，包括输入格式、算法实现及输出解析，适用于初学者理解字符串操作及动态规划的应用。

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Pieces

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1355 Accepted Submission(s): 688

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

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/*
 * Author:  ****
 * Created Time:  2013/9/18 16:11:27
 * File Name: A.cpp
 * solve: A.cpp
 */
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<string>
#include<map>
#include<stack>
#include<set>
#include<iostream>
#include<vector>
#include<queue>
//ios_base::sync_with_stdio(false);
//#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;
#define sz(v) ((int)(v).size())
#define rep(i, a, b) for (int i = (a); i < (b); ++i)
#define repf(i, a, b) for (int i = (a); i <= (b); ++i)
#define repd(i, a, b) for (int i = (a); i >= (b); --i)
#define clr(x) memset(x,0,sizeof(x))
#define clrs( x , y ) memset(x,y,sizeof(x))
#define out(x) printf(#x" %d\n", x)
#define sqr(x) ((x) * (x))
typedef long long LL;

const int INF = 1000000000;
const double eps = 1e-8;
const int maxn = (1<<16);

int sgn(const double &x) {  return (x > eps) - (x < -eps); }

char str[20];
bool is_palindrome[maxn];
int dp[maxn];
int len;
void init()
{
    char buf[20];
    rep(i,0,1<<len)
    {
        int cnt = 0;
        rep(j,0,len)
        {
            if((1<<j) & i)
            {
                buf[cnt] = str[j];
                cnt++;
            }
        }
        int flag = 1;
        rep(j,0,cnt)
        {
            if(buf[j] != buf[cnt - 1 - j])
            {
                flag = 0;
                break;
            }
        }
        is_palindrome[i] = flag;
    }
    return ;
}

void d()
{
    dp[0] = 0;
    rep(i,1,1<<len)
    {
        dp[i] = INF;
        for(int sub = i;sub > 0;sub = (sub - 1)&i)
        {
            if(is_palindrome[sub])
                dp[i] = min(dp[i],dp[sub^i] + 1);
        }
    }
}
int main() 
{
    //freopen("in.txt","r",stdin);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",str);
        len = strlen(str);
        init();
        d();
        cout<<dp[(1<<len) - 1]<<endl;
    }

    return 0;
}