HDU 3091 - Necklace - [状压DP]

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3091

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 327680/327680 K (Java/Others)

Problem Description

One day , Partychen gets several beads , he wants to make these beads a necklace . But not every beads can link to each other, every bead should link to some particular bead(s). Now , Partychen wants to know how many kinds of necklace he can make.

 

Input

It consists of multi-case .
Every case start with
two integers N,M ( 1<=N<=18,M<=N*N )
The followed M lines contains
two integers a,b ( 1<=a,b<=N ) which means the ath bead and the bth bead
are able to be linked.

 

Output

An integer , which means the number of kinds that the
necklace could be.

 

Sample Input

3 3

1 2

1 3

2 3

 

Sample Output

2

题意：

给你N个珠子，这些珠子编号为1~N，然后给出可以连在一起的两个珠子的编号，求把他们全部串起来有多少种方案。

样例有两种项链穿法：

　　①：“1-2-3-1”（包含“2-3-1-2”，“3-1-2-3”这两种情况）；

　　②：“1-3-2-1”（包含“3-2-1-3”，“2-1-3-2”这两种情况）；

可以看出，珠子串出的项链呈环状，但是珠子有规定好的逆时针或者顺时针的顺序，不能翻面；

题解：

设 i 表示状态：i转化为二进制数后，第k位（从右往左数）为0，表示k号珠子还没穿上去；为1，就代表已经穿上去了；

设 j 代表当前状态下，最后一个穿上去的是j号珠子；

设dp[i][j]表示在(i,j)状态下的方案数；

AC代码：

 #include<cstdio>
 #include<cstring>
 #include<vector>
 using namespace std;
 typedef long long ll;
 int n,m;
 bool link[][];
 ll dp[<<][],ans;
 int main()
 {
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         memset(link,,sizeof(link));
         memset(dp,,sizeof(dp));
         for(int i=,u,v;i<=m;i++)
         {
             scanf("%d%d",&u,&v);
             link[v][u]=link[u][v]=;
         }
 
         int ed_state=(<<n)-;
         dp[][]=;
         for(int state=;state<=ed_state;state++)
         {
             for(int i=;i<=n;i++)
             {
                 if( (state&(<<(i-)))== || dp[state][i]== ) continue;
                 for(int j=;j<=n;j++)
                 {
                     if(!link[i][j]) continue;//这两颗珠子不能连在一起，跳过
                     if( state & (<<(j-)) ) continue;//这颗珠子已经在项链上，跳过
                     int next_state=state|(<<(j-));
                     dp[next_state][j]+=dp[state][i];
                     //printf("dp[%d][%d]=%I64d\n",next_state,next_bead,dp[next_state][next_bead]);
                 }
             }
         }
 
         ans=;
         for(int i=;i<=n;i++) if(link[i][]) ans+=dp[ed_state][i];
         printf("%I64d\n",ans);
     }
 }