解题思路

初识模拟退火，于是拿这道题练练手

我们指定没有连边的结点连了一条权值为inf的边，成为完全图，避免了不存在的生成树

这样枚举根，然后每一次枚举都跑若干遍模拟退火

初始状态就是一张菊花图，父亲都指向当前为根的结点

每次产生相近解只需要随机改变一个结点的父亲，但是这样并不能保证产生的关系任然是一棵树

考虑到最多只有12个结点，每次跑一遍并查集暴力判断，可以归结到常数复杂度

多试几个种子就过了

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>int n,m,x,y,v;
long long M[20][20];
const long long inf=0x3f3f3f3f3f3f;
int Fa[20];namespace SA{const double start_T=10000;const double delta_T=0.993;const double low_T=1e-12;long long calc(int now,int fa,int dep){long long ret=dep*M[fa][now];for (int i=1;i<=n;i++){if (Fa[i]!=now||i==Fa[i]) continue;ret+=calc(i,now,dep+1);}return ret;}namespace MFS{int B[20];inline void init(){for (int i=1;i<=n;i++) B[i]=i;}inline int find(int k){return (k==B[k])?(k):(B[k]=find(B[k]));}void merge(int a,int b){int fa=find(a),fb=find(b);if (fa==fb) return;B[fa]=fb;}inline bool same(int a,int b){return find(a)==find(b);}}bool allowed(){MFS::init();for (int i=1;i<=n;i++){if (i!=Fa[i]&&MFS::same(i,Fa[i])) return false;MFS::merge(i,Fa[i]);}return true;}long long ans;inline double rand_double(){return rand()/(double)RAND_MAX;}long long SA_main(int root){ans=calc(root,0,0);double T=start_T;while (T>low_T){int change,to;do{change=rand()%n+1,to=rand()%n+1;}while (change==to||change==root);int ec=Fa[change];Fa[change]=to;if (!allowed()){Fa[change]=ec;continue;}long long nxt=calc(root,0,0);if (nxt<ans||exp((ans-nxt)/T)>rand_double()) ans=nxt;else Fa[change]=ec;T*=delta_T;}return ans;}
}long long Ans=inf;int main(){srand(/**/);//种子被和谐了scanf("%d%d",&n,&m);if (n==1){printf("%d\n",0);return 0;}memset(M,0x3f,sizeof(M));for (int i=1;i<=m;i++){scanf("%d%d%d",&x,&y,&v);M[x][y]=M[y][x]=std::min(M[x][y],(long long)v);}for (int i=0;i<=n;i++)for (int j=0;j<=n;j++)M[i][j]=std::min(M[i][j],inf);for (int i=1;i<=n;i++){for (int j=1;j<=n;j++) Fa[j]=i;//初始状态菊花图Ans=std::min(Ans,SA::SA_main(i));Ans=std::min(Ans,SA::SA_main(i));Ans=std::min(Ans,SA::SA_main(i));Ans=std::min(Ans,SA::SA_main(i));Ans=std::min(Ans,SA::SA_main(i));Ans=std::min(Ans,SA::SA_main(i));Ans=std::min(Ans,SA::SA_main(i));Ans=std::min(Ans,SA::SA_main(i));Ans=std::min(Ans,SA::SA_main(i));Ans=std::min(Ans,SA::SA_main(i));//这里多做几次}printf("%lld\n",Ans);
}