Design the city
題意：求三個節點到他們最近公共父節點的距離和
思路：對於三個節點兩兩求到最近公共父節點的距離，然後加起來除以二就是答案
參考代碼：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
#define N 100010
#define NN 500010
struct Edge
{
int v,len,next;
} edge[N*2];
int tot,head[N];
void add(int a,int b,int c)
{
edge[tot].v=b;
edge[tot].len=c;
edge[tot].next=head[a];
head[a]=tot++;
edge[tot].v=a;
edge[tot].len=c;
edge[tot].next=head[b];
head[b]=tot++;
}
struct Query{
int v,index,next;
};
Query Q[NN];
int q_tot,q_head[N];
void add_ques(int a,int b,int index)
{
Q[q_tot].v=b;
Q[q_tot].index=index;
Q[q_tot].next=q_head[a];
q_head[a]=q_tot++;
Q[q_tot].v=a;
Q[q_tot].index=index;
Q[q_tot].next=q_head[b];
q_head[b]=q_tot++;
}
bool vis[N];
int pre[N],dis[N],ans[NN/2];
int n,q;
void init(){
for(int i=0;i<=n;i++){
pre[i]=i;
vis[i]=false;
}
tot=q_tot=0;
memset(q_head,-1,sizeof(q_head));
memset(head,-1,sizeof(head));
memset(ans,-1,sizeof(ans));
}
int find(int x){
return pre[x]==x?pre[x]:pre[x]=find(pre[x]);
}
void Union(int x,int y){
x=find(x);
y=find(y);
if(x!=y){
pre[y]=x;
}
}
void Tarjan(int cur,int val){
dis[cur]=val;
vis[cur]=true;
for(int j=head[cur]; j!=-1; j=edge[j].next)
{
int v=edge[j].v;
if(vis[v]) continue;
Tarjan(v,val+edge[j].len);
Union(cur,v);
}
for(int j=q_head[cur]; j!=-1; j=Q[j].next)
{
int v=Q[j].v;
if(!vis[v]) continue;
ans[Q[j].index]=dis[v]+dis[cur]-2*dis[find(v)];
}
}
int main(){
int T=0,m;
// scanf("%d",&T);
while(scanf("%d",&n)!=EOF){
int a,b,c;
init();
for(int i=0;i<n-1;i++){
scanf("%d%d%d",&a,&b,&c);
add(a,b,c);
}
scanf("%d",&q);
for(int i=0;i<q;i++){
scanf("%d%d%d",&a,&b,&c);
add_ques(a,b,3*i);
add_ques(a,c,3*i+1);
add_ques(b,c,3*i+2);
}
Tarjan(0,0);
if(T) printf("\n");
T=1;
for(int i=0;i<q;i++){
printf("%d\n",(ans[3*i]+ans[3*i+1]+ans[3*i+2])/2);
}
}
return 0;
}