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 }
