HDU 3091

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[20][20];
ll dp[1<<18][20],ans;
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(link,0,sizeof(link));
        memset(dp,0,sizeof(dp));
        for(int i=1,u,v;i<=m;i++)
        {
            scanf("%d%d",&u,&v);
            link[v][u]=link[u][v]=1;
        }

        int ed_state=(1<<n)-1;
        dp[1][1]=1;
        for(int state=1;state<=ed_state;state++)
        {
            for(int i=1;i<=n;i++)
            {
                if( (state&(1<<(i-1)))==0 || dp[state][i]==0 ) continue;
                for(int j=1;j<=n;j++)
                {
                    if(!link[i][j]) continue;//这两颗珠子不能连在一起，跳过
                    if( state & (1<<(j-1)) ) continue;//这颗珠子已经在项链上，跳过
                    int next_state=state|(1<<(j-1));
                    dp[next_state][j]+=dp[state][i];
                    //printf("dp[%d][%d]=%I64d
",next_state,next_bead,dp[next_state][next_bead]);
                }
            }
        }

        ans=0;
        for(int i=1;i<=n;i++) if(link[i][1]) ans+=dp[ed_state][i];
        printf("%I64d
",ans);
    }
}