HDU 2586

http://acm.hdu.edu.cn/showproblem.php?pid=2586

题意：求最近祖先节点的权值和

思路：LCA Tarjan算法

 #include <stdio.h>
 #include <string.h>
 #define maxn 40005

 int m,n,x[maxn],y[maxn],z[maxn],head[maxn*],pos,dist[maxn],f[maxn];
 bool vis[maxn];
 struct Edge{
     int to,val,next;
 }edge[maxn*];

 void add(int u,int v,int val)
 {
     edge[pos].to = v;
     edge[pos].val = val;
     edge[pos].next = head[u];
     head[u] = pos++;
 }

 int Find(int x)
 {
     if(f[x]!=x)
         return f[x] =Find(f[x]);
     return f[x];
 }

 void Tarjan(int u)
 {
     vis[u] = true;
     f[u] = u;
     for(int i = ;i<=n;i++)
     {
         if(x[i]==u&&vis[y[i]]) z[i] = Find(y[i]);
         if(y[i]==u&&vis[x[i]]) z[i] = Find(x[i]);
     }
     for(int i =head[u];i!=-;i=edge[i].next)
     {
         int v = edge[i].to;
         if(!vis[v])
         {
             dist[v] =dist[u]+edge[i].val;
             Tarjan(v);
             f[v] = u;
         }
     }
 }

 int main()
 {
     int t,a,b,c;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d%d",&m,&n);
         memset(head,-,sizeof(head));
         memset(vis,false,sizeof(vis));

         pos = ;
         for(int i = ;i<m;i++)
         {
             scanf("%d%d%d",&a,&b,&c);
             add(a,b,c);
             add(b,a,c);
             z[i] = ;
         }
         z[m] = ;
         for(int i = ; i<=n;i++)
         {
             scanf("%d%d",&x[i],&y[i]);  //分别存每一次的查询点。
         }
         dist[] = ;
         Tarjan();
         for(int i = ;i<=n;i++)
             printf("%d\n",dist[x[i]]+dist[y[i]]-*dist[z[i]]);
     }
     return ;
 }