bzoj1036: [ZJOI2008]树的统计Count

最新推荐文章于 2019-09-07 10:57:07 发布

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#树链剖分

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本文详细介绍了树链剖分的概念及其在解决树形结构数据问题中的应用，并结合线段树实现高效的区间查询与更新操作。通过具体代码示例展示了树链剖分的实现过程。

树链剖分

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
int a[30010];
int read () {
	char c = getchar();
	int re = 0;
	int f = 1;
	while(c > '9' || c < '0') {
		if(c == '-')
			f = -1;
		c = getchar();
	}
	while(c <= '9' && c >= '0') {
		re = re * 10 + c - '0';
		c = getchar();
	}
	return re * f;
}
int n,u,v,q,lca;
char s[10];
int depth[30010],sz[30010],in[30010],son[30010],fa[30010],top[30010],ind;

struct LINE_TREE {
	int sum,MAX;
}tr[120010];

struct edge {
	int v,next;
}e[60010];
int cnt = -1,head[30010];
void adde (int u,int v) {
	e[++cnt].v = v;
	e[cnt].next = head[u];
	head[u] = cnt;
	e[++cnt].v = u;
	e[cnt].next = head[v];
	head[v] = cnt;
}

void dfs_1 (int U) {
	for(int i = head[U];i != -1;i = e[i].next) {
		int V = e[i].v;
		if(fa[U] == V)
			continue;
		depth[V] = depth[U] + 1;
		fa[V] = U;
		dfs_1(V);
		if(sz[son[U]] < sz[V])	
			son[U] = V;
		sz[U] += sz[V];
	}
	sz[U] += 1;
}

void dfs_2 (int U,int tp) {
	top[U] = tp;
	in[U] = ++ind;
	if(son[U])
		dfs_2(son[U],tp);
	for(int i = head[U];i != -1;i = e[i].next) {
		int V = e[i].v;
		if(fa[U] == V || V == son[U])
			continue;
		dfs_2(V,V);
	}
}

void push_up (int m) {
	if(tr[m << 1].MAX > tr[m << 1 | 1].MAX)
		tr[m].MAX = tr[m << 1].MAX;
	else tr[m].MAX = tr[m << 1 | 1].MAX;
	tr[m].sum = tr[m << 1].sum + tr[m << 1 | 1].sum;
}

void build (int l,int r,int m) {
	if(l == r) {
		tr[m].sum = tr[m].MAX = a[l];
		return;
	}
	int mid = (l + r) / 2;
	build(l,mid,m << 1);
	build(mid + 1,r,m << 1 | 1);
	push_up(m);
}

void change (int goal,int l,int r,int m) {
	if(l == r && l == goal) {
		tr[m].sum = tr[m].MAX = v;
		return;
	}
	int mid = (l + r) / 2;
	if(goal <= mid)
		change(goal,l,mid,m << 1);
	else change(goal,mid + 1,r,m << 1 | 1);
	push_up(m);
}

int Max (int L,int R,int l,int r,int m) {
	if(L == l && R == r)
		return tr[m].MAX;
	int mid = (l + r) / 2;
	if(L > mid)
		return Max(L,R,mid + 1,r,m << 1 | 1);
	else {
		if(R <= mid)
			return Max(L,R,l,mid,m << 1);
		else return max(Max(L,mid,l,mid,m << 1),Max(mid + 1,R,mid + 1,r,m << 1 | 1));
	}
}

int Sum (int L,int R,int l,int r,int m) {
	if(L == l && R == r)
		return tr[m].sum;
	int mid = (l + r) / 2;
	if(L > mid)
		return Sum(L,R,mid + 1,r,m << 1 | 1);
	else {
		if(R <= mid)
			return Sum(L,R,l,mid,m << 1);
		else return Sum(L,mid,l,mid,m << 1) + Sum(mid + 1,R,mid + 1,r,m << 1 | 1);
	}
}

int main () {
	freopen("FAQ.in","r",stdin);
	freopen("FAQ.in","w",stdout);
	memset(head,-1,sizeof head);
	n = read();
	for(int i = 1;i <= n - 1;++i) {
		u = read();
		v = read();
		adde(u,v);
	}
	depth[1] = 1;
	dfs_1(1);
	dfs_2(1,1);
	for(int i = 1;i <= n;i++)
		a[in[i]] = read();
	build(1,ind,1);
	q = read();
	while(q--) {
		scanf("%s",s);
		u = read();
		v = read();
		if(s[0] == 'C')
			change(in[u],1,ind,1);
		else {
			if(s[1] == 'M') {
				int re = -0x3f3f3f3f;
				
				int tu = top[u];
				int tv = top[v];
				while(tu != tv) {
					if(depth[tu] < depth[tv]) {
						swap(u,v);
						swap(tu,tv);
					}
					re = max(re,Max(in[top[u]],in[u],1,ind,1));
					u = fa[tu];
					tu = top[u];
				}
				if(depth[u] > depth[v])
					swap(u,v);
				re = max(re,Max(in[u],in[v],1,ind,1));
				printf("%d\n",re);
				
			}
			else {
				int re = 0;
				int tu = top[u];
				int tv = top[v];
				while(tu != tv) {
					if(depth[tu] < depth[tv]) {
						swap(u,v);
						swap(tu,tv);
					}
					re += Sum(in[top[u]],in[u],1,ind,1);
					u = fa[tu];
					tu = top[u];
				}
				if(depth[u] > depth[v])
					swap(u,v);
				re += Sum(in[u],in[v],1,ind,1);
				printf("%d\n",re);
			}
		}
	}
	return 0;
}