hdu - 2586 How far away ？（最短路共同祖先问题）

 1 # include<stdio.h>
 2 # include<string.h>
 3 # define N 40005
 4 # define M 205
 5 struct node{
 6     int from,to,next,val;
 7 }edge[2*N];
 8 struct node1{
 9     int from,to,next,num;
10 }edge1[2*M];
11 int tol,head[N],head1[N],tol1,father[N],dis[N],LCA[M],n,m;
12 bool visit[N];
13 void add(int a,int b,int c)
14 {
15     edge[tol].from=a;edge[tol].to=b;edge[tol].next=head[a];edge[tol].val=c;head[a]=tol++;
16 }
17 void add1(int a,int b,int c)
18 {
19     edge1[tol1].from=a;edge1[tol1].to=b;edge1[tol1].next=head1[a];edge1[tol1].num=c;head1[a]=tol1++;
20 }
21 int find(int x)
22 {
23     if(x!=father[x])
24         father[x]=find(father[x]);
25     return father[x];
26 }
27 void tarjan(int u)
28 {
29     int j,v;
30     visit[u]=1;
31     father[u]=u;
32     //////////////////
33     for(j=head1[u];j!=-1;j=edge1[j].next)
34     {
35         v=edge1[j].to;
36         if(visit[v]) LCA[edge1[j].num]=find(v);
37     }
38     //////////////////
39     for(j=head[u];j!=-1;j=edge[j].next)
40     {
41         v=edge[j].to;
42         if(!visit[v]) 
43         {
44             dis[v]=dis[u]+edge[j].val;
45             tarjan(v);
46             father[v]=u;
47         }
48     }
49 }
50 int main()
51 {
52     int i,ncase,a,b,c;
53     scanf("%d",&ncase);
54     while(ncase--)
55     {
56         scanf("%d%d",&n,&m);
57         tol=0;
58         memset(head,-1,sizeof(head));
59         for(i=1;i<n;i++)
60         {
61             scanf("%d%d%d",&a,&b,&c);
62             add(a,b,c);
63             add(b,a,c);
64         }
65         memset(visit,0,sizeof(visit));
66         tol1=0;
67         memset(head1,-1,sizeof(head1));
68         for(i=1;i<=m;i++)
69         {
70             scanf("%d%d",&a,&b);
71             add1(a,b,i);
72             add1(b,a,i);
73         }
74         ///LCA是一种离线算法，所以刚开始需要把所有的询问都输入，然后用邻接表进行存储，i表示第i次询问
75         dis[1]=0;
76         tarjan(1);
77         for(i=0;i<tol1;i+=2)
78         {
79             a=edge1[i].from;
80             b=edge1[i].to;
81             c=edge1[i].num;
82             printf("%d\n",dis[a]+dis[b]-2*dis[LCA[c]]);
83         }
84     }
85     return 0;
86 }