Codeforces Round #425 (Div. 2) D.Misha, Grisha and Underground

我奇特的脑回路的做法就是
树链剖分 + 树状数组

树状数组是那种 区间修改，区间求和，还有回溯的

当我看到别人写的是lca，直接讨论时，感觉自己的智商收到了碾压。。。

#include<cmath>
#include<map>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 1e5+5;
#define MS(x,y) memset(x,y,sizeof(x))
#define MP(x, y) make_pair(x, y)
const int INF = 0x3f3f3f3f;

int n, q;
struct Node{
    int to,next;
}edge[N << 1];
int tot;
int head[N];
void addedge(int x,int y){
    edge[tot].to = y; edge[tot].next = head[x]; head[x] = tot ++;
}

int top[N],fa[N],son[N],deep[N],num[N],p[N],fp[N],pos;
void dfs1(int x,int pre,int dep){
    deep[x] = dep;
    fa[x] = pre;
    num[x] = 1;
    for(int i = head[x]; i != -1; i = edge[i].next){
        int y = edge[i].to;
        if(y == pre) continue;
        dfs1(y,x,dep + 1);
        num[x] += num[y];
        if(son[x] == -1 || num[y] > num[son[x]])
            son[x] = y;
    }
}
void dfs2(int x,int tp){
    top[x] = tp;
    p[x] = pos++;
    fp[p[x]] = x;
    if(son[x] == -1) return;
    dfs2(son[x],tp);
    for(int i = head[x] ; i != -1; i = edge[i].next){
        int y = edge[i].to;
        if(y != son[x] && y != fa[x])
            dfs2(y,y);
    }
}

ll tree1[N]; ll tree2[N];
void Add(ll tree[], int pos, int val) {
    for(int i = pos; i <= n; i += i&-i) {
        tree[i] += val;
    }
}
ll Sum(ll tree[], int pos) {
    if(pos == 0) return 0;
    ll ans = 0;
    for(int i = pos; i; i -= i&-i) {
        ans += tree[i];
    }
    return ans;
}
vector<pair<int, int> > Resume;
void Find(int x, int y) {
    int fx = top[x]; int fy = top[y];
    while(fx != fy) {
        if(deep[fx] < deep[fy]) {
            swap(fx, fy);
            swap(x, y);
        }
        Add(tree1, p[fx], 1);
        Add(tree1, p[x]+1, -1);
        Add(tree2, p[fx], p[fx]);
        Add(tree2, p[x]+1, -p[x]-1);

        Resume.push_back(MP(p[fx], -1));
        Resume.push_back(MP(p[x]+1, 1));
        Resume.push_back(MP(-p[fx], -p[fx]));
        Resume.push_back(MP(-p[x]-1, p[x]+1));

        x = fa[fx];
        fx = top[x];
    }
    if(deep[x] > deep[y]) swap(x, y);
    Add(tree1, p[x], 1);
    Add(tree1, p[y]+1, -1);
    Add(tree2, p[x], p[x]);
    Add(tree2, p[y]+1, -p[y]-1);

    Resume.push_back(MP(p[x], -1));
    Resume.push_back(MP(p[y]+1, 1));
    Resume.push_back(MP(-p[x], -p[x]));
    Resume.push_back(MP(-p[y]-1, p[y]+1));
}
ll Total(int x, int y) {
    ll sum = 0;
    int fx = top[x]; int fy = top[y];
    while(fx != fy) {
        if(deep[fx] < deep[fy]) {
            swap(fx, fy);
            swap(x, y);
        }
        sum += 1ll*(p[x]+1)*Sum(tree1, p[x]) - Sum(tree2, p[x]) - 1ll*(p[fx])*Sum(tree1, p[fx]-1) + Sum(tree2, p[fx]-1);
        x = fa[fx];
        fx = top[x];
    }
    if(deep[x] > deep[y]) swap(x, y);
    sum += 1ll*(p[y]+1)*Sum(tree1, p[y]) - Sum(tree2, p[y]) - 1ll*(p[x])*Sum(tree1, p[x]-1) + Sum(tree2, p[x]-1);

    return sum;
}

ll solve(int a,int b, int c,int d) {
    Resume.clear();
    Find(a, b);
//  for(int i = 1; i <= n; ++i) printf("%d ", Sum(i)); printf("\n");
    ll tt = Total(c, d);
//  printf("hh %d %d %d %d %d\n",a,b,c,d, tt);
    for(int i = 0; i < Resume.size(); ++i) {
        if(Resume[i].first > 0) Add(tree1, Resume[i].first, Resume[i].second);
        else                    Add(tree2, -Resume[i].first, Resume[i].second);
    }
    return tt;
}
int main() {
    while(~scanf("%d %d", &n, &q)) {
        MS(tree1, 0); MS(tree2, 0);
        memset(head, -1, sizeof(head));
        memset(son, -1, sizeof(son));
        tot = 0; pos = 1;

        for(int i = 2; i <= n; ++i) {
            int a; scanf("%d", &a);
            addedge(a, i); addedge(i, a);
        }

        dfs1(1, 1, 1);
        dfs2(1, 1);

//      for(int i = 1; i <= n; ++i) printf("%d ", p[i]); printf("\n");

        for(int i = 0; i < q; ++i) {
            int a, b, c; scanf("%d %d %d", &a, &b, &c);
            ll ans = max(max(solve(a,b, a,c), solve(a,b, b,c)), solve(a,c, b,c));
            printf("%lld\n", ans);
        }

    }
    return 0;
}