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最新推荐文章于 2020-10-20 20:23:46 发布

原创最新推荐文章于 2020-10-20 20:23:46 发布 · 724 阅读

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本文介绍了一个有趣的编程问题：如何用最少的步骤将一个字符串通过删除其中的回文子序列直至完全消除。文章提供了完整的状态压缩动态规划解决方案，并详细解释了算法实现。

Pieces

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

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

状态压缩DP

#include<cstdio>
#include<cstring>

#define Min(a,b) (a<b?a:b)

int cnt,n;
char ch[100];
int d[1<<16];

void DFS(int tmp,int x,int y)
{
	int i,j,a;
	for(i=x+1;i<y;i++)
	{
		if(tmp&1<<i)
		{
			a=tmp^1<<i;
			d[a]=Min(d[a],d[cnt]+1);
		}
		else
			continue;
		for(j=y-1;j>i;j--)
		{
			if(ch[i]==ch[j] && tmp&1<<i && tmp&1<<j)
			{
				
				
				a=a^1<<j;
				d[a]=Min(d[a],d[cnt]+1);
				DFS(a,i,j);
				a=a^1<<j;
			}
		}
	}
}

int main()
{
	int T;
	int i,j;
	int tmp;

	scanf("%d",&T);

	while(T--)
	{
		scanf("%s",ch);
		n=strlen(ch);
		memset(d,1,sizeof(d));

		d[(1<<n)-1]=0;
		for(cnt=(1<<n)-1;cnt>0;cnt--)
		{
			if(d[cnt]>=d[0]-1)
				continue;
			for(i=0;i<n;i++)
			{
				if(cnt&1<<i)
				{
					tmp=tmp=cnt^1<<i;
					d[tmp]=Min(d[tmp],d[cnt]+1);
				}
				else
					continue;
				for(j=i+1;j<n;j++)
					if(ch[i]==ch[j] && cnt&1<<j)
					{
						tmp=tmp^1<<j;
						d[tmp]=Min(d[tmp],d[cnt]+1);
						DFS(tmp,i,j);
						tmp=tmp^1<<j;
					}					
			}
		}
		printf("%d\n",d[0]);
	}

	return 0;
}