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 = (1<<29);
const int MAXN = 30;
const int MAXS = (1<<20);
int n,m,ans,end;
int w[MAXN][MAXN];
int d[MAXN],f[MAXS];

inline int getint()
{
    int w=0,q=0; char c=getchar();
    while((c<'0' || c>'9') && c!='-') c=getchar(); if(c=='-') q=1,c=getchar(); 
    while (c>='0' && c<='9') w=w*10+c-'0', c=getchar(); return q ? -w : w;
}

inline void work(){
    n=getint(); m=getint(); int x,y,z,now; if(m==0) { printf("0"); return ; }
    for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) w[i][j]=inf;
    for(int i=1;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=1;k<=n;k++) for(int i=1;i<=n;i++) if(i!=k) for(int j=1;j<=n;j++) if(i!=j && j!=k) w[i][j]=min(w[i][j],w[i][k]+w[k][j]);
    for(int i=2;i<=n;i++) if(d[i]!=0 && w[1][i]==inf) { printf("-1"); return ; }
    for(int i=1;i<=n;i++) if(d[i]&1) end|=(1<<(i-1));
    for(int i=0;i<=end;i++) f[i]=inf; f[end]=0;
    for(int i=end;i>0;i--) {
    if(f[i]==inf) continue;
    for(int j=1;j<=n;j++) {
        if(((1<<(j-1))&i )==0) continue;
        for(int k=1;k<=n;k++) {
        if(( (1<<(k-1))&i )==0) continue; now=i^(1<<(k-1))^(1<<(j-1));
        f[now]=min(f[now],f[i]+w[j][k]);
        }
    }
    }
    ans+=f[0]; printf("%d",ans);
}

int main()
{
    work();
    return 0;
}