codeforces 21D：Traveling Graph

Description

You are given undirected weighted graph. Find the length of the shortest cycle which starts from the vertex 1 and passes throught all the edges at least once. Graph may contain multiply edges between a pair of vertices and loops (edges from the vertex to itself).

Input

The first line of the input contains two integers n and m (1 ≤ n ≤ 15, 0 ≤ m ≤ 2000), n is the amount of vertices, and m is the amount of edges. Following m lines contain edges as a triples x, y, w (1 ≤ x, y ≤ n, 1 ≤ w ≤ 10000), x, y are edge endpoints, and w is the edge length.

Output

Output minimal cycle length or -1 if it doesn't exists.

Examples

Input

3 3
1 2 1
2 3 1
3 1 1

Output

3

Input

3 2
1 2 3
2 3 4

Output

14
 
正解：状压DP
解题报告：
　　考虑度数为偶数的点不需要被考虑，只需要考虑度数为奇数的情况。首先每条边必须要访问一次，所以所有边权加起来就是答案的初值。　
　　然后度数为奇数的点就需要访问之前已经走过的边。我们考虑两个度数为奇数的点可以组合一下，变成度数为偶数的点，相当于是在这两个点之间新连了一条边，我们可以floyd预处理一下两点之间的最短路。然后状压，状态表示当前哪些结点的度数为奇数，然后枚举每次连哪两条边就可以了。
　　

 //It is made by jump~
 #include <iostream>
 #include <cstdlib>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <ctime>
 #include <vector>
 #include <queue>
 #include <map>
 #include <set>
 using namespace std;
 typedef long long LL;
 const int inf = (<<);
 const int MAXN = ;
 const int MAXS = (<<);
 int n,m,ans,end;
 int w[MAXN][MAXN];
 int d[MAXN],f[MAXS];
 
 inline int getint()
 {
     int w=,q=; char c=getchar();
     while((c<'' || c>'') && c!='-') c=getchar(); if(c=='-') q=,c=getchar();
     while (c>='' && c<='') w=w*+c-'', c=getchar(); return q ? -w : w;
 }
 
 inline void work(){
     n=getint(); m=getint(); int x,y,z,now; if(m==) { printf(""); return ; }
     for(int i=;i<=n;i++) for(int j=;j<=n;j++) w[i][j]=inf;
     for(int i=;i<=m;i++) {
     x=getint(); y=getint(); z=getint();
     ans+=z; d[x]++; d[y]++;
     if(w[x][y]>z) w[x][y]=z,w[y][x]=z;
     }
     for(int k=;k<=n;k++) for(int i=;i<=n;i++) if(i!=k) for(int j=;j<=n;j++) if(i!=j && j!=k) w[i][j]=min(w[i][j],w[i][k]+w[k][j]);
     for(int i=;i<=n;i++) if(d[i]!= && w[][i]==inf) { printf("-1"); return ; }
     for(int i=;i<=n;i++) if(d[i]&) end|=(<<(i-));
     for(int i=;i<=end;i++) f[i]=inf; f[end]=;
     for(int i=end;i>;i--) {
     if(f[i]==inf) continue;
     for(int j=;j<=n;j++) {
         if(((<<(j-))&i )==) continue;
         for(int k=;k<=n;k++) {
         if(( (<<(k-))&i )==) continue; now=i^(<<(k-))^(<<(j-));
         f[now]=min(f[now],f[i]+w[j][k]);
         }
     }
     }
     ans+=f[]; printf("%d",ans);
 }
 
 int main()
 {
     work();
     return ;
 }