ZOJ 3396

斯坦那树有一个结论：n个点的斯坦那树至少会经过n-2个中间节点，这里n=3，所以至少会经过一个中间节点，求出这三个点的单源最短路后枚举中间节点即可。

要注意虽然枚举有一些点时，并不是简单的三个dist相加，但最小的一定是三个相加

#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#define N 510#define inf 6000000using namespace std;struct Edge{    int v,w,next;}edge[12000*2];int val[11000],head[N],cnt,dis[N][N];bool vis[N],ok[10100];void init(int n){    int i,j;    memset(head,-1,sizeof(head));    memset(ok,0,sizeof(ok));    for(i=1;i<=n;i++)        for(j=1;j<=n;j++)            dis[i][j]=inf;    cnt=0;}void addedge(int u,int v,int w){    edge[cnt].v=v;    edge[cnt].w=w;    edge[cnt].next=head[u];    head[u]=cnt++;    edge[cnt].v=u;    edge[cnt].w=w;    edge[cnt].next=head[v];    head[v]=cnt++;}void spfa(int s){    int i;    queue<int>q;    dis[s][s]=0;    memset(vis,0,sizeof(vis));    q.push(s);    vis[s]=1;    while(!q.empty()){        int u=q.front();        q.pop();        vis[u]=0;        for(i=head[u];i!=-1;i=edge[i].next){            int v=edge[i].v;            if(dis[s][v]>dis[s][u]+edge[i].w){                dis[s][v]=dis[s][u]+edge[i].w;                if(!vis[v]){                    vis[v]=1;                    q.push(v);                }            }        }    }}int main(){    int n,m,l,i,j,q;    int u,v,w;    int x,y,z;    int ans;    int cas=0;    while(scanf("%d %d %d",&n,&m,&l)==3){        init(m);        for(i=1;i<=n;i++)            scanf("%d",&val[i]);        for(i=1;i<=l;i++){            scanf("%d %d %d",&u,&v,&w);            addedge(u,v,w);        }        scanf("%d",&q);        printf("Case #%d\n",++cas);        for(j=1;j<=q;j++){            scanf("%d %d %d",&x,&y,&z);            if(ok[val[x]]==0){                spfa(val[x]);                ok[val[x]]=1;            }            if(ok[val[y]]==0){                spfa(val[y]);                ok[val[y]]=1;            }            if(ok[val[z]]==0){                spfa(val[z]);                ok[val[z]]=1;            }            ans=inf*3;            for(i=1;i<=m;i++)                if(dis[val[x]][i]!=inf && dis[val[y]][i]!=inf && dis[val[z]][i]!=inf)                    ans=min(ans,dis[val[x]][i]+dis[val[y]][i]+dis[val[z]][i]);            if(ans==inf*3)                printf("Line %d: Impossible to connect!\n",j);            else                printf("Line %d: The minimum cost for this line is %d.\n",j,ans);        }    }}