Geiz-Revive > 洛谷 P4556 [Vani有约会] 雨天的尾巴

洛谷 P4556 [Vani有约会] 雨天的尾巴

2019-06-04

线段树合并

线段树合并入门题（然而还是因为$sb$错误交了好几次），考虑每个节点建一棵动态开点的线段树，每次进行树上差分，这里给出一些主要代码片段：

更新：

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inl void maintain(int x) {
	if (t[ls(x)].w >= t[rs(x)].w) {
		t[x].w = t[ls(x)].w, t[x].c = t[ls(x)].c;
	}
	else t[x].w = t[rs(x)].w, t[x].c = t[rs(x)].c;//c表示颜色，w表示权值
}

合并：

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inl int merge(int p, int q, int l, int r) {
	if (!p || !q)return p + q;
	if (l == r) {
		t[p].w += t[q].w, t[p].c = t[q].c;
		return p;
	}
	re mid = l + r >> 1;
	ls(p) = merge(ls(p), ls(q), l, mid);
	rs(p) = merge(rs(p), rs(q), mid + 1, r);
	maintain(p);
	return p;
}

还有一个比较迷的点就是当我没有离散化时

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inl void maintain(int x) {
	if (t[ls(x)].w >= t[rs(x)].w) {
		t[x].w = t[ls(x)].w, t[x].c = t[ls(x)].c;
	}
	else t[x].w = t[rs(x)].w, t[x].c = t[rs(x)].c;
    if(!t[x].w)t[x].c = 0;
}
inl void solve(int x) {
	for (re i = head[x]; i; i = e[i].next) {
		if (e[i].to != fa[x]) {
			solve(e[i].to);
			root[x] = merge(root[x], root[e[i].to], 1, s);
		}
	}
	ans[x] = t[root[x]].w;
}

这样写是对的

但是如果离散化以后，这样就只有95$pts$???

于是乎经多次测试，离散化以后应该这样写：

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inl void maintain(int x) {
	if (t[ls(x)].w >= t[rs(x)].w) {
		t[x].w = t[ls(x)].w, t[x].c = t[ls(x)].c;
	}
	else t[x].w = t[rs(x)].w, t[x].c = t[rs(x)].c;
}
inl void solve(int x) {
	for (re i = head[x]; i; i = e[i].next) {
		if (e[i].to != fa[x]) {
			solve(e[i].to);
			root[x] = merge(root[x], root[e[i].to], 1, s);
		}
	}
	ans[x] = t[root[x]].w ? mp[t[root[x]].c] : 0;
}

我认为这两种写法应该是完全没有差别的，但是结果并不一样$emmm$。

全部代码：

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#pragma region revive
#include <set>
#include <map>
#include <cmath>
#include <queue>
#include <stack>
#include <bitset>
#include <cstdio>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <unordered_map>
#define inl inline
#define re register int
#define fa(x) t[x].fa
#define son(x,y) t[x].child[y]
#define ls(x) t[x].child[0]
#define rs(x) t[x].child[1]
#define ll long long
const int inf = 0x3f3f3f3f;
#define lowbit(x) ((x) & (-x))
using namespace std;
#ifndef _DEBUG
#define getchar() (*(IOB.in.p++))
#define putchar(c) (*(IOB.out.p++)=(c))
#define io_eof() (IOB.in.p>=IOB.in.pend)
struct IOBUF { struct { char buff[1 << 26], *p, *pend; }in; struct { char buff[1 << 26], *p; }out; IOBUF() { in.p = in.buff; out.p = out.buff; in.pend = in.buff + fread(in.buff, 1, 1 << 26, stdin); }~IOBUF() { fwrite(out.buff, 1, out.p - out.buff, stdout); } }IOB;
#endif
template<typename IO>
inl void write(IO x) {
	if (x == 0) return (void)putchar('0');
	if (x < 0)putchar('-'), x = -x;
	static char buf[30];
	char* p = buf;
	while (x) {
		*(p++) = x % 10 + '0';
		x /= 10;
	}
	while (p > buf)putchar(*(--p));
}
inl void writestr(const char *s) { while (*s != 0)putchar(*(s++)); }
template<typename IO>inl void writeln(IO x) { write(x), putchar('\n'); }
template<typename IO>inl void writesp(IO x) { write(x), putchar(' '); }
inl int readstr(char *s) {
	char *begin = s, c = getchar();
	while (c < 33 || c>127) {
		c = getchar();
	}
	while (c >= 33 && c <= 127) {
		*(s++) = c;
		c = getchar();
	}
	*s = 0;
	return s - begin;
}
template<typename IO>
inl IO read() {
	IO x = 0;
	register bool w = 0;
	register char c = getchar();
	while (c > '9' || c < '0') {
		if (c == '-') w = 1;
		c = getchar();
	}
	while (c >= '0' && c <= '9') {
		x = (x << 3) + (x << 1) + (c ^ 48);
		c = getchar();
	}
	return w ? -x : x;
}
#pragma endregion
#undef ls
#undef rs
#define ls(x) t[x].l
#define rs(x) t[x].r
int cnt, s;
struct node {
	int l, r, w, c;
}t[10000001];
inl void maintain(int x) {
	if (t[ls(x)].w >= t[rs(x)].w) {
		t[x].w = t[ls(x)].w, t[x].c = t[ls(x)].c;
	}
	else t[x].w = t[rs(x)].w, t[x].c = t[rs(x)].c;
}
inl void add(int &k, int l, int r, int p, int w) {
	if (!k)k = ++cnt;
	if (l == r) {
		t[k].w += w, t[k].c = p;
		return;
	}
	re mid = l + r >> 1;
	if (p <= mid)add(ls(k), l, mid, p, w);
	else add(rs(k), mid + 1, r, p, w);
	maintain(k);
}
inl int merge(int p, int q, int l, int r) {
	if (!p || !q)return p + q;
	if (l == r) {
		t[p].w += t[q].w, t[p].c = t[q].c;
		return p;
	}
	re mid = l + r >> 1;
	ls(p) = merge(ls(p), ls(q), l, mid);
	rs(p) = merge(rs(p), rs(q), mid + 1, r);
	maintain(p);
	return p;
}
int head[1000001], tot, top[1000001], dep[1000001], h[1000001], fa[1000001], siz[1000001], root[1000001], a[1000001], mp[1000001], ans[1000001];
struct edge {
	int next, to;
}e[1000001];
inl void adde(int x, int y) {
	e[++tot] = edge{ head[x],y }, head[x] = tot;
	e[++tot] = edge{ head[y],x }, head[y] = tot;
}
inl void dfs1(int x) {
	dep[x] = dep[fa[x]] + 1, siz[x] = 1;
	for (re i = head[x]; i; i = e[i].next) {
		if (fa[x] != e[i].to) {
			fa[e[i].to] = x;
			dfs1(e[i].to);
			siz[x] += siz[e[i].to];
			siz[e[i].to] > siz[h[x]] ? h[x] = e[i].to : 0;
		}
	}
}
inl void dfs2(int x) {
	if (h[x]) {
		top[h[x]] = top[x];
		dfs2(h[x]);
	}
	for (re i = head[x]; i; i = e[i].next) {
		if (!top[e[i].to]) {
			top[e[i].to] = e[i].to;
			dfs2(e[i].to);
		}
	}
}
inl int lca(int x, int y) {
	while (top[x] != top[y]) {
		if (dep[top[x]] < dep[top[y]])swap(x, y);
		x = fa[top[x]];
	}
	return dep[x] < dep[y] ? x : y;
}
inl void solve(int x) {
	for (re i = head[x]; i; i = e[i].next) {
		if (e[i].to != fa[x]) {
			solve(e[i].to);
			root[x] = merge(root[x], root[e[i].to], 1, s);
		}
	}
	ans[x] = t[root[x]].w ? mp[t[root[x]].c] : 0;
}
struct quiz {
	int x, y, z;
}q[1000001];
signed main() {
	re n = read<int>(), m = read<int>(), x, y, z, f;
	top[1] = 1;
	for (re i = 1; i < n; i++)adde(read<int>(), read<int>());
	for (re i = 1; i <= m; i++) x = read<int>(), y = read<int>(), a[i] = z = read<int>(), q[i] = quiz{ x,y,z };
	sort(a + 1, a + 1 + m);
	s = unique(a + 1, a + 1 + m) - a - 1;
	dfs1(1), dfs2(1);
	for (re i = 1, k; i <= m; i++) k = q[i].z, mp[q[i].z = lower_bound(a + 1, a + 1 + s, q[i].z) - a] = k;
	for (re i = 1; i <= m; i++) {
		x = q[i].x, y = q[i].y, z = q[i].z, f = lca(x, y);
		add(root[x], 1, s, z, 1), add(root[y], 1, s, z, 1), add(root[f], 1, s, z, -1);
		if (fa[f])add(root[fa[f]], 1, s, z, -1);
	}
	solve(1);
	for (re i = 1; i <= n; i++)writeln(ans[i]);
}