【BZOJ 3626】【LNOI 2014】LCA

BZOJ 3626

题意

给你一棵树，定义根节点的深度为1。
有若干询问，每次询问l r z，求出[l,r]所有节点和z节点的公共祖先深度之和。（模201314）

样例输入

5 2
0
0
1
1
1 4 3
1 4 2

样例输出

8
5

SOL

我真的很想吐槽这道题作为一道省选题没有部分分！
所以这道题目是不是只有0分和100分QAQ
首先考虑一下一组操作怎么做。首先你要想到差分，也就是说[l,r]和z求答案等价于[1,r]的值减去[1,l]的值，虽然这并没有什么帮助。。。
然后考虑一种奇奇怪怪的做法：把z到根节点的路径上全部打上标记，那么任何一个节点到根的路径上有几个标记，lca的深度就是多少。
这个操作是可逆的，也就是说我们可以对[1,l]这l个点到根的路径上打上标记，然后算一下z到根的路径上有多少标记。
想到这里差不多就做出来了。每次取出一个节点，把这个节点到根的路径加1，如果恰好需要询问这一组解，那么就求出z到根的节点权值和。

#include<cmath>#include<cstdio>#include<vector>#include<cstring>#include<iomanip>#include<stdlib.h>#include<iostream>#include<algorithm>#define ll long long#define inf 1000000000#define mod 1000000007#define N 300005using namespace std;struct tree2 {int len,siz; ll sum;} tree[N*4];vector<int> v[N],ask[N];int dep[N],siz[N],fa[N],id[N],son[N],val[N],top[N];int num,n,qq,i,j,x,y,t,f;int z[N];ll lazy[N*4],res[N];void dfs1(int x,int ftr,int d){    dep[x] = d; siz[x] = 1; son[x] = 0; fa[x] = ftr;    for (int i = 0;i < v[x].size(); i++)        {            int nxt = v[x][i];            if (nxt == ftr) continue;            dfs1(nxt,x,d+1);            siz[x] += siz[nxt];            if (siz[nxt] > siz[son[x]]) son[x] = nxt;        }}void dfs2(int x,int tp){    top[x] = tp; id[x] = ++num;    if (son[x] > 0) dfs2(son[x],tp);    for (int i = 0;i < v[x].size(); i++)        {            int nxt = v[x][i];            if (nxt == fa[x] || nxt == son[x]) continue;            dfs2(nxt,nxt);        }}void build(int l,int r,int rt){    tree[rt].len = r - l + 1; tree[rt].sum = 0;    if (l == r) {tree[rt].siz = 1; return;}    int mid = (l + r) >> 1;    build(l,mid,rt << 1);    build(mid+1,r,rt << 1 | 1);    tree[rt].siz = tree[rt<<1].siz + tree[rt<<1|1].siz + 1;}void pushdown(int x){    if (lazy[x] == 0) return;    ll t = lazy[x];    lazy[x<<1] += t; lazy[x<<1|1] += t;    tree[x<<1].sum += tree[x<<1].len * t;    tree[x<<1|1].sum += tree[x<<1|1].len * t;    lazy[x] = 0;}void update(int rt,int l,int r,int L,int R,int s){    tree[rt].sum = tree[rt].sum % 201314;    if (L <= l && r <= R)         {            lazy[rt] += (ll) s;            tree[rt].sum += (ll) s * tree[rt].len;            return;        }    pushdown(rt);    int mid = (l + r) >> 1;    if (L <= mid) update(rt<<1,l,mid,L,R,s);    if (mid < R) update(rt<<1|1,mid+1,r,L,R,s);    tree[rt].sum = tree[rt<<1].sum + tree[rt<<1|1].sum;}ll query(int rt,int l,int r,int L,int R){    pushdown(rt);    if (L <= l && r <= R) return tree[rt].sum;    int mid = (l + r) >> 1;    ll res = 0;    if (L <= mid) res += query(rt<<1,l,mid,L,R);    if (mid < R) res += query(rt<<1|1,mid+1,r,L,R);    return res;}ll link(int u,int v,int opt){    int top1 = top[u]; int top2 = top[v];    ll res = 0;    while (top1 != top2)        {            if (dep[top1] < dep[top2])                {swap(top1,top2); swap(u,v);}            if (opt == 1) update(1,1,n,id[top1],id[u],1);            if (opt == 2) res += query(1,1,n,id[top1],id[u]);            res = res % 201314;            u = fa[top1];            top1 = top[u];        }    if (dep[u] > dep[v]) swap(u,v);    if (opt == 1) update(1,1,n,id[u],id[v],1);    if (opt == 2) res += query(1,1,n,id[u],id[v]);    return res % 201314;}int main(){    scanf("%d%d",&n,&qq);     for (i = 2;i <= n; i++)        {            scanf("%d",&x); x++;            v[i].push_back(x); v[x].push_back(i);           }    num = 0; dfs1(1,0,1); dfs2(1,1); build(1,n,1);    for (i = 1;i <= qq; i++)        {            scanf("%d%d%d",&x,&y,&z[i]); x++; y++; z[i]++;            ask[x-1].push_back(-i);            ask[y].push_back(i);        }    for (i = 1;i <= n; i++)        {            link(i,1,1);            for (j = 0;j < ask[i].size(); j++)                {                    t = ask[i][j];                    if (t < 0) {t = - t; f = -1;} else f = 1;                    res[t] += f * link(z[t],1,2);                }        }    for (i = 1;i <= qq; i++) printf("%lld\n",(res[i] + 2013140) % 201314);    return 0;}