（状压dp）ABC 067 F : Mole and Abandoned Mine

Mole decided to live in an abandoned mine. The structure of the mine is represented by a simple connected undirected graph which consists of N vertices numbered 1through N and M edges. The i-th edge connects Vertices a_i and b_i, and it costs c_iyen (the currency of Japan) to remove it.

Mole would like to remove some of the edges so that there is exactly one path from Vertex 1 to Vertex N that does not visit the same vertex more than once. Find the minimum budget needed to achieve this.

Constraints

• 2≤N≤15
• N?1≤M≤N(N?1)?2
• 1≤a_i,b_i≤N
• 1≤c_i≤10⁶
• There are neither multiple edges nor self-loops in the given graph.
• The given graph is connected.

Input

Input is given from Standard Input in the following format:

N M
a1 b1 c1
:
aM bM cM

Output

Print the answer.

Sample Input 1

4 6
1 2 100
3 1 100
2 4 100
4 3 100
1 4 100
3 2 100

Sample Output 1

200

By removing the two edges represented by the red dotted lines in the figure below, the objective can be achieved for a cost of 200 yen.

Sample Input 2

2 1
1 2 1

Sample Output 2

0

It is possible that there is already only one path from Vertex 1 to Vertex N in the beginning.

Sample Input 3

15 22
8 13 33418
14 15 55849
7 10 15207
4 6 64328
6 9 86902
15 7 46978
8 14 53526
1 2 8720
14 12 37748
8 3 61543
6 5 32425
4 11 20932
3 12 55123
8 2 45333
9 12 77796
3 9 71922
12 15 70793
2 4 25485
11 6 1436
2 7 81563
7 11 97843
3 1 40491

Sample Output 3

133677

同时进行两种递推：1、加入单独1点 2、加入一与当前无交集的点集　　　　get： line50 如何取得补集的所有子集。

 1 #include <bits/stdc++.h>
 2 typedef long long ll;
 3 using namespace std;
 4 const int MAX=1e5+5;
 5 const int INF=1e9;
 6 int n,m;
 7 int edge[20][20];
 8 int total;
 9 int cost[1<<16][16],dp[1<<16][16];//dp[s][x]记录达到某点集s,最后“末端”（继续连接其余点的唯一点）为x时余下权值和最大的边的情况
10 int inner[1<<16];//内部的边
11 int main()
12 {
13     scanf("%d%d",&n,&m);
14     for(int i=1;i<=m;i++)//读入数据
15     {
16         int x,y,price;
17         scanf("%d%d%d",&x,&y,&price);
18         --x;--y;
19         edge[x][y]=edge[y][x]=price;
20     }
21     total=1<<n;
22     for(int i=0;i<total;i++)//计算某点集内部所有边权值之和
23         for(int a=0;a<n;a++)
24             if(i&(1<<a))
25                 for(int b=0;b<a;b++)
26                     if(i&(1<<b))
27                         if(edge[a][b])
28                             inner[i]+=edge[a][b];
29     for(int i=0;i<total;i++)//计算某点集到点集外某点所有边的权值之和
30         for(int a=0;a<n;a++)
31             if(!(i&(1<<a)))
32                 for(int b=0;b<n;b++)
33                     if((i&(1<<b))&&edge[a][b])
34                         cost[i][a]+=edge[a][b];
35     memset(dp,-1,sizeof(dp));
36     dp[1][0]=0;//只有1个点的集合dp值显然为0
37     for(int i=0;i<total;i++)
38     {
39
40         for(int a=0;a<n;a++)
41         {
42             if((i&(1<<a))&&dp[i][a]!=-1)
43             {
44                 for(int b=0;b<n;b++)
45                     if(!(i&(1<<b)))/*只加入一个点*/
46                         if(edge[a][b])
47                             dp[i|(1<<b)][b]=max(dp[i|(1<<b)][b],dp[i][a]+edge[a][b]);
48                 /*加入一个点集*/
49                 int left=total-1-i;
50                 for(int now=left;now!=0;now=(now-1)&left)//用此循环得到所有
51                 {
52                     int num=inner[now]+cost[now][a];
53                     dp[i|now][a]=max(dp[i|now][a],dp[i][a]+num);
54                 }
55             }
56         }
57     }
58     printf("%d\n",inner[total-1]-dp[total-1][n-1]);
59     return 0;
60 }