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;

}