luogu4211 LCA

题目链接

思路

我们换一种求\(dep[lca(i,j)]\)的方法。

将从根到\(i\)的路径上所有点的权值加\(1\)，然后求从根节点到j路径上点的权值和。就是\(i\)和\(j\)的\(lca\)的深度。

以此类推，对于求\(\sum\limits_{i=l}^rdep[lca(i,z)]\)，我们可以对于从l到r中的每个节点到根节点的路径上的点权值加\(1\)，然后求一边从\(z\)到根节点路径和即可。这个可以用树链剖分做到。

然后再考虑多次询问。发现这个题可以离线。那就比较好处理了。

因为每次询问都是形如\([l,r]\)的形式。所以可以通过\([0,r]-[0,l-1]\)来得到\([l,r]\)的答案。所以我们对于询问离线之后，分别在每次询问的\(l-1\)和\(r\)处询问一次。相减就是答案。

代码

/*
* @Author: wxyww
* @Date:   2019-01-29 14:50:20
* @Last Modified time: 2019-01-29 16:13:25
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<bitset>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
#define int ll
const int N = 100010,mod = 201314;
vector<int>e[N];
ll read() {
	ll x=0,f=1;char c=getchar();
	while(c<'0'||c>'9') {
		if(c=='-') f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9') {
		x=x*10+c-'0';
		c=getchar();
	}
	return x*f;
}

int dep[N],top[N],son[N],dfn[N],siz[N],fa[N];
void dfs1(int u,int father) {
	fa[u] = father;dep[u] = dep[father] + 1;siz[u] = 1;
	int k = e[u].size();
	for(int i = 0;i < k;++i) {
		int v = e[u][i];
		if(v == father) continue;
		dfs1(v,u);
		siz[u] += siz[v];
		if(siz[v] > siz[son[u]]) son[u] = v;
	}
}

int tot;
void dfs2(int u,int father,int tt) {
	dfn[u] = ++tot;top[u] = tt;
	int k = e[u].size();
	if(!son[u]) return;
	dfs2(son[u],u,tt);
	for(int i = 0;i < k;++i) {
		int v = e[u][i];
		if(v == father || v == son[u]) continue;
		dfs2(v,u,v);
	}
}

int tree[N << 2],lazy[N << 2];
void pushup(int rt) {
	tree[rt] = (tree[rt << 1] + tree[rt << 1 | 1]) % mod;
}

void pushdown(int rt,int ln,int rn) {
	if(lazy[rt]) {
		lazy[rt << 1] += lazy[rt];
		lazy[rt << 1] %= mod;
		lazy[rt << 1 | 1] += lazy[rt];
		lazy[rt << 1 | 1] %= mod;
		tree[rt << 1] += lazy[rt] * ln;
		tree[rt << 1] %= mod;
		tree[rt << 1 | 1] += lazy[rt] * rn;
		tree[rt << 1 | 1] %= mod;
		lazy[rt] = 0;
	}
}

void update(int rt,int l,int r,int L,int R,int c) {
	if(L <= l && R >= r) {
		lazy[rt] += c;
		lazy[rt] %= mod;
		tree[rt] += c * (r - l + 1);
		tree[rt] %= mod;
		return;
	}
	int mid = (l + r) >> 1;
	pushdown(rt,mid - l + 1,r - mid);
	if(L <= mid) update(rt << 1,l,mid,L,R,c);
	if(R > mid) update(rt << 1 | 1,mid + 1,r,L,R,c);
	pushup(rt);
}

int query(int rt,int l,int r,int L,int R) {
	if(L <= l && R >= r)
		return tree[rt];
	int mid = (l + r) >> 1;
	pushdown(rt,mid - l + 1,r - mid);
	int ans = 0;
	if(L <= mid) ans += query(rt << 1,l,mid,L,R),ans %= mod;
	if(R > mid) ans += query(rt << 1 | 1,mid + 1,r,L,R),ans %= mod;
	return ans;
}

void UPDATE(int rt,int c) {
	while(rt != 0) {
		update(1,1,tot,dfn[top[rt]],dfn[rt],c);
		rt = fa[top[rt]];
	}
}

int Query(int rt) {
	int ans = 0;
	while(rt != 0) {
		ans += query(1,1,tot,dfn[top[rt]],dfn[rt]);
		ans %= mod;
		rt = fa[top[rt]];
	}
	return ans;
}

struct node {
	int id,z,x,ans;
}ask[N];

bool cmp(node x,node y) {
	return x.x < y.x;
}

bool cmp2(node x,node y) {
	return x.id < y.id;
}
int js;
signed main() {
	int n = read(),Q = read();
	for(int i = 2;i <= n;++i) {
		int x = read() + 1;
		e[i].push_back(x);e[x].push_back(i);
	}
	for(int i = 1;i <= Q;++i) {
		int l = read() + 1,r = read() + 1,z = read() + 1;
		ask[++js].id = js;
		ask[js].x = l - 1;ask[js].z = z;
		ask[++js].id = js;
		ask[js].x = r;ask[js].z = z;
	}
	dfs1(1,0);
	dfs2(1,0,1);
	sort(ask + 1,ask + js + 1,cmp);
	int now = 1;
	while(ask[now].x == 0) ++now;
	for(int i = 1;i <= n;++i) {
		UPDATE(i,1);
		while(ask[now].x == i) {
			ask[now].ans = Query(ask[now].z);
			++now;
		}
	}
	sort(ask + 1,ask + js + 1,cmp2);
	for(int i = 2;i <= js;i += 2)
		printf("%lld\n",(ask[i].ans - ask[i - 1].ans + mod) % mod);
	return 0;
}