ZOJ Problem Set - 3195 Design the city 【Tarjan离线LCA】

题目：ZOJ Problem Set - 3195 Design the city

题意：给出一个图，求三点的连起来的距离。

分析：分别求出三点中任意两点的距离 / 2  = ans

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
#define N 50010
#define M 20010

struct Node
{
    int to,val;
};
vector<Node> g[N];
struct ask{
    int u,v,lca;
};
vector<ask> e;
vector<int> query[N];
int fa[N],ance[N],dir[N];
bool vis[N];

inline void add_Node(int u ,int v,int w)
{
    g[u].push_back((Node){v,w});
    g[v].push_back((Node){u,w});
}

inline void add_ask(int u ,int v )
{
    e.push_back((ask){u,v,-1});
    e.push_back((ask){v,u,-1});
    int len = e.size()-1;
    query[v].push_back(len);
    query[u].push_back(len-1);
}

int find(int x){
    return x == fa[x] ? x : fa[x] = find(fa[x]);
}

void Tarjan(int u,int val)
{
    vis[u] = true;
    ance[u] = fa[u] = u;
    dir[u] = val;
    for(int i=0;i<g[u].size();i++)
    {
        Node tmp = g[u][i];
        if(!vis[tmp.to])
        {
            Tarjan(tmp.to,val+tmp.val);
            fa[tmp.to] = u;
        }
    }
    for(int i=0;i<query[u].size();i++)
    {
        int num = query[u][i];
        ask& tmp = e[num];
        if(vis[tmp.v])
        {
            tmp.lca = e[num^1].lca = ance[find(tmp.v)];
        }
    }
}
void Clear(int n)
{
    memset(vis,false,sizeof(vis));
    for(int i=0;i<=n;i++){
        g[i].clear();
        query[i].clear();
    }
    e.clear();
}
int main()
{
    //freopen("Input.txt","r",stdin);
    int n,m,fack = 0;
    while(~scanf("%d",&n))
    {
        Clear(n);
        for(int i=1;i<n;i++)
        {
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            add_Node(x,y,z);
        }
        scanf("%d",&m);
        for(int i=0;i<m;i++)
        {
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            add_ask(x,y);
            add_ask(x,z);
            add_ask(y,z);
        }
        Tarjan(0,0);
        int ans = 0;
        if(fack)
            puts("");
        fack = 1;
        for(int i=0;i<3*m;i++)
        {
            ans+=( dir[e[i*2].u] + dir[e[i*2].v] - 2*dir[e[i*2].lca] );
            if(i%3==2){
                printf("%d\n",ans/2);
                ans = 0;
            }
        }
    }
    return 0;
}