（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 ai and bi, and it costs ciyen (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≤ai,bi≤N
  • 1≤ci≤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.

45c15676bb602ca3b762561fc014ecd0.png

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 }

 

posted @ 2017-07-17 19:08  perplex  阅读(425)  评论(0)  编辑  收藏  举报