（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。

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
const int MAX=1e5+5;
const int INF=1e9;
int n,m;
int edge[20][20];
int total;
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
int inner[1<<16];//内部的边
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)//读入数据
    {
        int x,y,price;
        scanf("%d%d%d",&x,&y,&price);
        --x;--y;
        edge[x][y]=edge[y][x]=price;
    }
    total=1<<n;
    for(int i=0;i<total;i++)//计算某点集内部所有边权值之和
        for(int a=0;a<n;a++)
            if(i&(1<<a))
                for(int b=0;b<a;b++)
                    if(i&(1<<b))
                        if(edge[a][b])
                            inner[i]+=edge[a][b];
    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
        for(int a=0;a<n;a++)
            if(!(i&(1<<a)))
                for(int b=0;b<n;b++)
                    if((i&(1<<b))&&edge[a][b])
                        cost[i][a]+=edge[a][b];
    memset(dp,-1,sizeof(dp));
    dp[1][0]=0;//只有1个点的集合dp值显然为0
    for(int i=0;i<total;i++)
    {

        for(int a=0;a<n;a++)
        {
            if((i&(1<<a))&&dp[i][a]!=-1)
            {
                for(int b=0;b<n;b++)
                    if(!(i&(1<<b)))/*只加入一个点*/
                        if(edge[a][b])
                            dp[i|(1<<b)][b]=max(dp[i|(1<<b)][b],dp[i][a]+edge[a][b]);
                /*加入一个点集*/
                int left=total-1-i;
                for(int now=left;now!=0;now=(now-1)&left)//用此循环得到所有
                {
                    int num=inner[now]+cost[now][a];
                    dp[i|now][a]=max(dp[i|now][a],dp[i][a]+num);
                }
            }
        }
    }
    printf("%d
",inner[total-1]-dp[total-1][n-1]);
    return 0;
}