洛谷p4211 [LNOI2014] LCA

[LNOI2014] LCA

题面

洛谷

题解

水题。
对于一个区间\([l,r]\)的答案。
我们把它差分成\([1,r] - [1,l - 1]\)
然后因为右端点是固定的,用树剖+只有1个端点的莫队就解决了...
你谷恶评实在太严重了。

#include <bits/stdc++.h>

const int maxn = 5e4 + 10;
const int mod = 201314;    

template<class t> inline void read(t& res) {
	res = 0;  char ch = getchar();  bool neg = 0;
	while (!isdigit(ch))
		neg |= ch == '-', ch = getchar();
	while (isdigit(ch))
		res = (res << 1) + (res << 3) + (ch & 15), ch = getchar();
	if (neg)
		res = -res;    
}

int q, n, m, i, j, k, cntq;     
int hd[maxn], ver[maxn << 1], nxt[maxn << 1], ans[maxn];        
struct Questions {
	int p, z, id; int type;  
	Questions() { p = z = type = id = 0; }
	Questions(int _p, int _z, int i, int _t) { p = _p;  z = _z;  id = i;  type = _t; }
	inline friend bool operator < (Questions a, Questions b) { return a.p < b.p; }
} Q[maxn << 1];

inline void adde(int u, int v) {
	static int cnte = 0;
	ver[++cnte] = v;  nxt[cnte] = hd[u];
	hd[u] = cnte;  return;
} 

int s[maxn << 2], tag[maxn << 2];   
inline void push_up(int u) { s[u] = (s[u << 1] + s[u << 1 | 1]) % mod; }     
inline void down(int u, int l, int r, int v) { s[u] += (r - l + 1) * v;  tag[u] += v; }
inline void push_down(int u, int l, int r) {
	int mid = (l + r) >> 1;
	down(u << 1, l, mid, tag[u]);
	down(u << 1 | 1, mid + 1, r, tag[u]);
	tag[u] = 0;   
}
void segt_modify(int ml, int mr, int l, int r, int u, int v) {     
//	printf("%d %d %d %d %d %d\n", ml, mr, l, r, u, v);      
	if (ml <= l && r <= mr)
		return s[u] += (r - l + 1) * v, tag[u] += v, void();
	int mid = (l + r) >> 1;
	push_down(u, l, r);
	if (ml <= mid)
		segt_modify(ml, mr, l, mid, u << 1, v);
	if (mid < mr)
		segt_modify(ml, mr, mid + 1, r, u << 1 | 1, v);
	push_up(u);
}
int segt_query(int ql, int qr, int l, int r, int u) {
	if (ql <= l && r <= qr)
		return s[u];
	int mid = (l + r) >> 1, res = 0;
	push_down(u, l, r);
	if (ql <= mid)
		res += segt_query(ql, qr, l, mid, u << 1) % mod;
	res %= mod;  
	if (mid < qr)
		res += segt_query(ql, qr, mid + 1, r, u << 1 | 1) % mod;
	res %= mod;   
	return res;    
}

namespace Tree_Chain {
int hsn[maxn], sz[maxn], dep[maxn], fa[maxn], top[maxn];
int dfn[maxn], tim;      
void dfs1(int u, int pa) {
	dep[u] = dep[pa] + 1;
	fa[u] = pa;
	sz[u] = 1;
	for (int i = hd[u]; ~i; i = nxt[i]) {
		int v = ver[i];
		if (v == pa)
			continue;
		dfs1(v, u);
		sz[u] += sz[v];
		if (sz[v] > sz[ hsn[u] ])
			hsn[u] = v;
	}
}
void dfs2(int u, int up) {
	top[u] = up;  dfn[u] = ++tim;
	if (hsn[u])
		dfs2(hsn[u], up);     
	for (int i = hd[u]; ~i; i = nxt[i]) {
		int v = ver[i];
		if (v == fa[u] || v == hsn[u])
			continue;
		dfs2(v, v);     
	}
}
void modify(int u, int v, int val) {
	while (top[u] != top[v]) {
		if(dep[ top[u] ] < dep[ top[v] ])
			std::swap(u, v);  
		segt_modify(dfn[ top[u] ], dfn[u], 1, n, 1, val);
		u = fa[ top[u] ];
	}
	if (dep[u] > dep[v])
		std::swap(u, v);
	segt_modify(dfn[u], dfn[v], 1, n, 1, val);  
}
int query(int u, int v) {
	int res = 0;  
	while (top[u] != top[v]) {
		//printf("%d %d\n", u, v);
		if (dep[ top[u] ] < dep[ top[v] ])
			std::swap(u, v);
		res += segt_query(dfn[ top[u] ], dfn[u], 1, n, 1) % mod;   
		res %= mod;
		u = fa[ top[u] ];
	}
	if (dep[u] > dep[v])
		std::swap(u, v);
	res += segt_query(dfn[u], dfn[v], 1, n, 1) % mod;
	res %= mod;
	return res;    
}
}

int main() {
	memset(hd, -1, sizeof(hd));  
	read(n);  read(q);
	for (int i = 2, p; i <= n; i++)
		read(p), p++, adde(p, i), adde(i, p);
	Tree_Chain::dfs1(1, 0);   
	Tree_Chain::dfs2(1, 1);
	for (int i = 1, l, r, z; i <= q; i++) {
		read(l);  read(r);  read(z);  z++;      
		Q[++cntq] = Questions(l, z, i, -1);
		Q[++cntq] = Questions(r + 1, z, i, 1);   
	}
	std::sort(Q + 1, Q + cntq + 1);
	int p = 1;
	for (int i = 1; i <= cntq; i++) {
		while (p <= Q[i].p)
			Tree_Chain::modify(1, p++, 1);  
		ans[ Q[i].id ] += Tree_Chain::query(1, Q[i].z) * Q[i].type;  
	}
	for (int i = 1; i <= q; i++)
		printf("%d\n", (ans[i] + mod) % mod);
	return 0;    
}
posted @ 2019-08-20 13:11  _connect  阅读(92)  评论(0编辑  收藏
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