HDU4628+状态压缩DP

/*
状态压缩DP
题意：每次可以去掉一个回文串，求最少几步能取完。
*/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<math.h>
using namespace std;
typedef long long ll;
//typedef __int64 int64;
const int maxn = 18;
const int inf = 0x3f3f3f3f;
const double pi=acos(-1.0);
const double eps = 1e-8;
int dp[ 1<<maxn ];
char s[ maxn ];
int state[ 1<<maxn ];//回文的状态

bool JudgeOneZero( int ss,int len ){
	int Index[ maxn ];
	int cc = 0;
	int IndexOfString = 0;
	while( IndexOfString<len ){
		if( ss%2==1 ){// 所有的 “1” 代表该位置上有字母，即这些组合是回文串
			Index[ cc++ ] = IndexOfString;
		}
		ss /= 2;
		IndexOfString++;
	}
	if( cc==1 ) return true;
	int L,R;
	L = 0;
	R = cc-1;
	while( L<=R ){
		if( s[Index[L]]!=s[Index[R]] ) return false;
		L++;
		R--;
	}
	return true;
}

int init_state( int len ){
	int cnt = 0;
	int N = 1<<len;
	state[ cnt++ ] = 0;
	for( int i=1;i<N;i++ ){
		if( JudgeOneZero( i,len )==true ){
			state[ cnt++ ] = i;
		}
	}
	return cnt;
} //初始化

bool Judge( int cur,int nxt,int len ){//当前状态cur，前一状态nxt
	int Index[ maxn ];
	int cc = 0;
	int IndexOfString = 0;
	while( IndexOfString<len ){
		if( cur%2==1 ){
			if( nxt%2==0 ) return false;
		}//当前状态为1，前一状态必须为1
		if( nxt%2==0 ){
			if( cur%2==1 ) return false;
		}//前一状态是0，当前状态也必须是0
		if( cur%2==0&&nxt%2==1 ){
			Index[ cc++ ] = IndexOfString;
		}
		IndexOfString++; 
		cur /= 2;
		nxt /= 2;
	}
	if( cc==1 ) return true;
	int L,R;
	L = 0;
	R = cc-1;
	//printf("cc=%d\n",cc);
	while( L<=R ){
		if( s[Index[L]]!=s[Index[R]] ) return false;
		L++;
		R--;
	}
	return true;
}

int main(){
	int T;
	scanf("%d",&T);
	while( T-- ){
		scanf("%s",s);
		int n = strlen(s);
		int cnt = init_state( n );
		int N = (1<<n);
		for( int i=0;i<N;i++ )
			dp[ i ] = inf;
		dp[ N-1 ] = 0;
		/*
		for( int i=N-2;i>=0;i-- ){
			 for( int j=0;j<N;j++ ){
			 	if( i==j ) continue;
			 	if( Judge( i,j,n )==true ){
			 		//printf("i=%d, j=%d\n",i,j);
			 		dp[ i ] = min( dp[i],dp[j]+1 );
			 		//printf("dp[%d] = %d\n\n",i,dp[i]);
			 	}
			 }
		}
		*/
		for( int i=N-2;i>=0;i-- ){
			for( int j=0;j<cnt;j++ ){
				if( 0==(i&state[j]) ){
					dp[ i ] = min( dp[i],dp[state[j]|i]+1 );
				}
			}
		}			
		printf("%d\n",dp[0]);
	}
	return 0;
}

补充：软件开发 , C++ ,