HDU-4628 Pieces 搜索 | DP

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4628

　　数据不大，枚举本质。首先对枚举出回文串，然后用DP或者搜索，这里因为层数不多，用bfs比较好，或者用IDA*。。。

  1 //STATUS:C++_AC_140MS_780KB

  2 #include <functional>

  3 #include <algorithm>

  4 #include <iostream>

  5 //#include <ext/rope>

  6 #include <fstream>

  7 #include <sstream>

  8 #include <iomanip>

  9 #include <numeric>

 10 #include <cstring>

 11 #include <cassert>

 12 #include <cstdio>

 13 #include <string>

 14 #include <vector>

 15 #include <bitset>

 16 #include <queue>

 17 #include <stack>

 18 #include <cmath>

 19 #include <ctime>

 20 #include <list>

 21 #include <set>

 22 #include <map>

 23 using namespace std;

 24 //using namespace __gnu_cxx;

 25 //define

 26 #define pii pair<int,int>

 27 #define mem(a,b) memset(a,b,sizeof(a))

 28 #define lson l,mid,rt<<1

 29 #define rson mid+1,r,rt<<1|1

 30 #define PI acos(-1.0)

 31 //typedef

 32 typedef __int64 LL;

 33 typedef unsigned __int64 ULL;

 34 //const

 35 const int N=(1<<16)+10;

 36 const int INF=0x3f3f3f3f;

 37 const int MOD=100000,STA=8000010;

 38 const LL LNF=1LL<<60;

 39 const double EPS=1e-8;

 40 const double OO=1e15;

 41 const int dx[4]={-1,0,1,0};

 42 const int dy[4]={0,1,0,-1};

 43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

 44 //Daily Use ...

 45 inline int sign(double x){return (x>EPS)-(x<-EPS);}

 46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}

 47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}

 48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}

 49 template<class T> inline T Min(T a,T b){return a<b?a:b;}

 50 template<class T> inline T Max(T a,T b){return a>b?a:b;}

 51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}

 52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}

 53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}

 54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}

 55 //End

 56 

 57 char s[18];

 58 int sta[N],d[N],t[18];

 59 int T,len,k,up;

 60 

 61 bool check(int status)

 62 {

 63     int i,j;

 64     for(i=j=0;status;i++,status>>=1)

 65         if(status&1)t[j++]=i;

 66     for(i=0;i<(j>>1);i++)

 67         if(s[t[i]]!=s[t[j-i-1]])return false;

 68     return true;

 69 }

 70 

 71 void getsta()

 72 {

 73     int i;

 74     up=(1<<(len=strlen(s)))-1;

 75     k=0;

 76     for(i=1;i<=up;i++)

 77         if(check(i))sta[k++]=i;

 78 }

 79 

 80 int bfs()

 81 {

 82     int i,u,v;

 83     queue<int> q;

 84     q.push(0);

 85     mem(d,-1);d[0]=0;

 86     while(!q.empty()){

 87         u=q.front();q.pop();

 88         for(i=0;i<k;i++){

 89             v=sta[i];

 90             if((u&v)==0 && d[u|v]==-1){

 91                 d[u|v]=d[u]+1;

 92                 q.push(u|v);

 93                 if((u|v)==up)return d[u|v];

 94             }

 95         }

 96     }

 97     return -1;

 98 }

 99 

100 int main()

101 {

102   //  freopen("in.txt","r",stdin);

103     int i,j;

104     scanf("%d",&T);

105     while(T--)

106     {

107         scanf("%s",s);

108         getsta();

109         printf("%d\n",bfs());

110     }

111     return 0;

112 }