hdu4628之状态压缩dp

Pieces

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1518    Accepted Submission(s): 788

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
   
   
   
   
   
   
   
   


   
   
   
   

分析：对于字符串len个位置,假设在某个字串含有原串k位置则k位置表示为1,否则表示为0，则枚举所有字串共有2^len种

对于每种字串状态i,枚举包含状态i的状态j(既i中有1的位置j也有)，然后判断状态j表示的字串消除的串i^j是否是回文串，是得话就可以从状态j到达状态i

对于如何枚举包含状态i的状态j:

for(int j=i;j<2^len;j=(j+1)|i);

比如:

i:1 1 0 1 0

j;1 1 0 1 0

则j+1:1 1 0 1然后(j+1)|i就将i中第一个为0的位置变为1

然后继续(j+1)|i其实就是在原前面已变位置的前提下，如果该位置前面还有0的就变成1，否则找下一个位置变为1

当然也可以枚举状态i的子状态:

for(int j=i;j>0;j=(j-1)&i);

枚举包含状态i的状态:

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <iomanip>
#define INF 99999999
typedef long long LL;
using namespace std;

const int MAX=(1<<16)+10;
int len,bit;
int dp[MAX];
char s[20];
bool mark[MAX];

bool check(int x){
	if(x == 0)return true;
	int i=0,j=len-1;
	while(i<j){
		while(!(x&(1<<i)))++i;
		while(!(x&(1<<j)))--j;
		if(s[i] != s[j])return false;
		++i,--j;
	}
	return true;
}

int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		scanf("%s",s);
		len=strlen(s);
		bit=1<<len;
		dp[bit-1]=0;
		for(int i=0;i<bit;++i)mark[i]=check(i);
		for(int i=bit-2;i>=0;--i){
			dp[i]=INF;
			for(int j = i;j<bit;j=(j+1)|i){
				if(!mark[i^j])continue;
				if(dp[i]>dp[j]+1)dp[i]=dp[j]+1;
			}
		}
		printf("%d\n",dp[0]);
	}
	return 0;
}

枚举状态i的子状态：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <iomanip>
#define INF 99999999
typedef long long LL;
using namespace std;

const int MAX=(1<<16)+10;
int len,bit;
int dp[MAX];
char s[20];
bool mark[MAX];

bool check(int x){
	if(x == 0)return true;
	int i=0,j=len-1;
	while(i<j){
		while(!(x&(1<<i)))++i;
		while(!(x&(1<<j)))--j;
		if(s[i] != s[j])return false;
		++i,--j;
	}
	return true;
}

int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		scanf("%s",s);
		len=strlen(s);
		bit=1<<len;
		for(int i=0;i<bit;++i)mark[i]=check(i);
		for(int i=0;i<bit;++i)dp[i]=INF;
		dp[bit-1]=0;
		for(int i=bit-1;i>=0;--i){
			for(int j = i;j>0;j=((j-1)&i)){
				if(!mark[i^j])continue;
				if(dp[j]>dp[i]+1)dp[j]=dp[i]+1;
			}
			if(mark[i])if(dp[0]>dp[i]+1)dp[0]=dp[i]+1;
		}
		printf("%d\n",dp[0]);
	}
	return 0;
}