指针版lca之小机房的树

题目>>http://codevs.cn/problem/2370/

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;const int MAXN = 4e5 + 5;int n,m,firs[MAXN],nex[MAXN],cnt = 0,num = 0,a,b,c;struct zt{    int f,t,v;}road[MAXN];struct zz{    int num,deep,dis;//num:该数字.deep:深度.dis:到根节点的距离.     zz *fa[20];//第2的几次方的父亲的地址. }tree[MAXN];zz *fin[MAXN];//某数字的地址. void build(int f,int t,int v){    road[++cnt] = (zt){f,t,v};    nex[cnt] = firs[f];    firs[f] = cnt;}void build_tree(zz *root,int x,int f,int v){    root->num = x;root->dis = v;fin[x] = root;    root->deep = root->fa[0]->deep + 1;//初始化     for(int i = firs[x];i != -1;i = nex[i])    {        int u = road[i].t;        if(u != f)//只能由父亲走向儿子.         {            tree[++num].fa[0] = root;//建立关系.             build_tree(&tree[num],u,x,v + road[i].v);//取地址.         }    }}void dfs(){    for(int i = 1;i < log(n) + 1;i ++)        for(int j = 0;j < n;j ++)//二进制记录其他祖先             fin[j]->fa[i] = fin[j]->fa[i - 1]->fa[i - 1];}int lca(int a,int b){    if(fin[a]->deep < fin[b]->deep)        swap(a,b);    for(int i = log(n);i >= 0;i --)//先跳到同一深度         if(fin[a]->fa[i]->deep >= fin[b]->deep)            a = fin[a]->fa[i]->num;    if(a == b) return a;//若是同一个点则之前a与b是"父子"关系.     for(int i = log(n);i >= 0;i --)//否则找他们的公共父亲的两个儿子.         if(fin[a]->fa[i] != fin[b]->fa[i])            a = fin[a]->fa[i]->num,b = fin[b]->fa[i]->num;    return fin[a]->fa[0]->num;//返回他们的父亲. }int main(){    scanf("%d",&n);    for(int i = 0;i <= n;i ++) firs[i] = -1;    for(int i = 1;i < n;i ++)    {        scanf("%d%d%d",&a,&b,&c);        build(a,b,c);        build(b,a,c);    }//以上为基本建图     tree[0].fa[0] = &tree[0];//确立根节点的父亲     build_tree(tree,0,0,0);    dfs();    scanf("%d",&m);    for(int i = 1;i <= m;i ++)    {        scanf("%d%d",&a,&b);        cout<<(fin[a]->dis + fin[b]->dis - (fin[lca(a,b)]->dis << 1))<<'\n';    }}