BZOJ3590 SNOI2013Quare（状压dp）

　　可能作为最优解的边双都可以这样生成：初始时边双内只有一个点，每次选取边双内部两点（可以相同）和一个当前不在边双内的点集，以该两点为起止点找一条链（当然如果两点相同就是个环）将点集串起来，加入边双。状压dp模拟这个过程即可。注意找链时对二元环特判。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 12
#define M 42
#define inf 100000000
#define rep(i,t,S) for (int t=S,i=lg2[t&-t];t;t^=t&-t,i=lg2[t&-t])
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,m,a[N][N],f[1<<N],g[N][N][1<<N],lg2[1<<N];
struct data{int x,y,z;
}edge[M];
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj3590.in","r",stdin);
    freopen("bzoj3590.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    int T=read();
    while (T--)
    {
        n=read(),m=read();
        for (int i=1;i<=m;i++) edge[i].x=read()-1,edge[i].y=read()-1,edge[i].z=read();
        memset(f,42,sizeof(f));for (int i=0;i<n;i++) f[1<<i]=0,lg2[1<<i]=i;
        memset(g,42,sizeof(g));
        for (int i=0;i<n;i++)
            for (int j=0;j<n;j++)
                for (int k=1;k<=m;k++)
                if (edge[k].x==i&&edge[k].y==j||edge[k].x==j&&edge[k].y==i)
                g[i][j][0]=min(g[i][j][0],edge[k].z);
        for (int i=1;i<(1<<n);i++)
        {
            rep(x,p,(1<<n)-1^i) rep(y,q,(1<<n)-1^i)
            if (x!=y||i!=(i&-i)) rep(j,o,i) g[x][y][i]=min(g[x][y][i],g[j][y][i^(1<<j)]+g[j][x][0]);
            else
            {
                int mn=inf,mn2=inf;
                for (int j=1;j<=m;j++)
                if (edge[j].x==x&&edge[j].y==lg2[i]||edge[j].x==lg2[i]&&edge[j].y==x)
                if (edge[j].z<mn) mn2=mn,mn=edge[j].z;
                else if (edge[j].z<mn2) mn2=edge[j].z;
                g[x][y][i]=mn+mn2;
            }
        }
        for (int i=1;i<(1<<n);i++)
            for (int j=i-1&i;j;j=j-1&i)
                rep(x,u,i^j) rep(y,v,i^j)
                {
                    f[i]=min(f[i],f[i^j]+g[x][y][j]);
                    if (x==y) break;
                }
        if (f[(1<<n)-1]<inf) printf("%d\n",f[(1<<n)-1]);
        else printf("impossible\n");
    }
    return 0;
}