SDOI_2014旅行

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SDOI_2014旅行

Solution

这道题考察了两个知识点，一个是树链剖分，另一个是动态开点线段树。

树剖不会的同学请点击这

我们着重来讲一下动态开点线段树

首先我们知道要开一个普通的线段树至少需要元素个数的$4$倍，对于这道题，如果对每一个宗教都开一个线段树显然不行，我们要对线段树进行改变，只保留树上有用的节点，单独用两个数组存左右儿子，查询时，遇到空区就直接返回，下面是两个图，可以帮助大家理解($Drawn$ $by$ $zyb$)

修改9

再修改6

但这种线段树的空间再不超出限制的情况下尽量开最大，否则会$\mathrm{RE}$或$\mathrm{WA}$

修改函数两个，一个是把原来的去掉，另一个是加上新的内容，请读者根据$code$自行理解。。。

Code

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#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define gc getchar()
#define pc(x) putchar(x)
#define re register
const int Maxn = 1e5 + 10;
inline int sc() {
    int xx = 0, ff = 1; char cch = gc;
    while(!isdigit(cch)) {
        if(cch == '-') ff = -1; cch = gc;
    }
    while(isdigit(cch)) {
        xx = (xx << 1) + (xx << 3) + (cch ^ '0'); cch = gc;
    }
    return xx * ff;
}
inline void out(int x) {
    if(x < 0) pc('-'), x = -x;
    if(x >= 10)
        out(x / 10);
    pc(x % 10 + '0');
}
struct node {
	int next, to;
}t[Maxn << 1];
struct LST {
	int sum, maxx, l, r;
}lst[Maxn * 40];
int n, m, cnt, len;
int w[Maxn], c[Maxn], head[Maxn], root[Maxn];
int siz[Maxn], top[Maxn], fa[Maxn], dep[Maxn], son[Maxn], id[Maxn];
inline void ADD(int from, int to) {
	t[++cnt].next = head[from];
	t[cnt].to = to;
	head[from] = cnt;
}
inline void dfs1(int now, int Fa, int Dep) {
	fa[now] = Fa;
	dep[now] = Dep;
	siz[now] = 1;
	for(re int i = head[now]; i; i = t[i].next) {
		int to = t[i].to;
		if(to == Fa) continue;
		dfs1(to, now, Dep + 1);
		siz[now] += siz[to];
		if(siz[son[now]] < siz[to])
			son[now] = to;
	}
}
inline void dfs2(int now, int topf) {
	id[now] = ++cnt;
//	rid[cnt] = now;
	top[now] = topf;
	if(!son[now]) return;
	dfs2(son[now], topf);
	for(re int i = head[now]; i; i = t[i].next) {
		int to = t[i].to;
		if(to == son[now] || to == fa[now]) continue;
		dfs2(to, to);
	}
}
inline void modify(int &k, int l, int r, int x, int z) {
	if(!k) k = ++len;
	lst[k].maxx = std :: max(lst[k].maxx, z), lst[k].sum += z;
	if(l == r) return;
	int mid = l + r >> 1;
	if(x <= mid) modify(lst[k].l, l, mid, x, z);
	else modify(lst[k].r, mid + 1, r, x, z);
}
inline void push(int &k, int l, int r, int x) {
	if(l == r) {
		lst[k].maxx = lst[k].sum = 0;
		return;
	}
	int mid = l + r >> 1;
	if(x <= mid) push(lst[k].l, l, mid, x);
	else push(lst[k].r, mid + 1, r, x);
	lst[k].maxx = std :: max(lst[lst[k].l].maxx, lst[lst[k].r].maxx);
	lst[k].sum = lst[lst[k].l].sum + lst[lst[k].r].sum;
}
inline int querymax(int k, int l, int r, int x, int y) {
	if(x > r || l > y) return 0;
	if(x <= l && r <= y) {
		return lst[k].maxx;
	}
	int mid = l + r >> 1;
	return std :: max(querymax(lst[k].l, l, mid, x, y), querymax(lst[k].r, mid + 1, r, x, y));
}
inline int querysum(int k, int l, int r, int x, int y) {
	if(x > r || l > y) return 0;  //y < l || x > r
	if(x <= l && r <= y) { //y >= r && x <= l  xlry
		return lst[k].sum;
	}
	int mid = l + r >> 1;
	return querysum(lst[k].l, l, mid, x, y) + querysum(lst[k].r, mid + 1, r, x, y);
}
inline int qymax(int x, int y, int cc) {
	int ans = 0;
	while(top[x] != top[y]) {
		if(dep[top[x]] < dep[top[y]]) std :: swap(x, y);
		ans = std :: max(ans, querymax(root[cc], 1, n, id[top[x]], id[x]));
		x = fa[top[x]];
	}
	if(dep[x] > dep[y]) std :: swap(x, y);
	ans = std :: max(ans, querymax(root[cc], 1, n, id[x], id[y]));
	return ans;
}
inline int qysum(int x, int y, int cc) {
	int ans = 0;
	while(top[x] != top[y]) {
		if(dep[top[x]] < dep[top[y]]) std :: swap(x, y);
		ans += querysum(root[cc], 1, n, id[top[x]], id[x]);
		x = fa[top[x]];
	}
	if(dep[x] > dep[y]) std :: swap(x, y);
	ans += querysum(root[cc], 1, n, id[x], id[y]);
	return ans;
}
int main() {
	n = sc(), m = sc();
	for(re int i = 1; i <= n; ++i)
		w[i] = sc(), c[i] = sc();
	for(re int i = 1; i < n; ++i) {
		int x = sc(), y = sc();
		ADD(x, y), ADD(y, x);
	}
	cnt = 0;
	dfs1(1, 0, 1), dfs2(1, 1);
	for(re int i = 1; i <= n; ++i)
		modify(root[c[i]], 1, n, id[i], w[i]);
	while(m--) {
		char ccc[5];
		std :: cin >> ccc;
		int x = sc(), y = sc();
		if(ccc[0] == 'C') {
			if(ccc[1] == 'C') {
//				std :: cerr << root[c[x]] << std :: endl;
				push(root[c[x]], 1, n, id[x]);
				modify(root[y], 1, n, id[x], w[x]);
				c[x] = y;
			}
			else {
//				std :: cerr << root[c[x]] << std :: endl;
				push(root[c[x]], 1, n, id[x]);
				modify(root[c[x]], 1, n, id[x], y);
				w[x] = y;
			}
		}
		else {
			if(ccc[1] == 'S') {
				out(qysum(x, y, c[x])), pc('\n');
			}
			else {
				out(qymax(x, y, c[x])), pc('\n');
			}
		}
	}
    return 0;
}
// Coded by dy.

rp++