BZOJ 3083 遥远的国度 树链剖分

标签：bzoj   树链剖分   子树查询   

题目大意：给出一颗无根树，有链的修改操作，还有子树的查询。除此之外，还有选定这棵树的一个点为根。

思路：子树操作，链上修改，带size域的树链剖分就可以搞定。换根肯定不能真的换，出题人要是闲的没事所有操作都在换根就惨。我们可以画一张图模拟下换根。先按照读入的顺序建一颗有根树，然后观察当前的根在要询问的点的位置。如果当前的根在要询问的点的儿子中，那么那个点为根的时候，当前点的子树就是除了当前点的有根节点的子树的点意外的所有点。如果根不在当前点的子树中，那么当前点的子树不变。还有一种情况要特殊讨论一下，当前点就是根，那么子树就是整棵树。

PS：一个点的子树的范围是pos[x]~pos[x] + size[x] - 1

CODE：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 100010
#define INF 0x3f3f3f3f
#define LEFT (pos << 1)
#define RIGHT (pos << 1|1)
using namespace std;

struct SegmentTree{
	int _min;
	bool v;
}tree[MAX << 2];

int points,asks;
int head[MAX],total;
int next[MAX << 1],aim[MAX << 1];

int deep[MAX],father[MAX],size[MAX],son[MAX];
int pos[MAX],top[MAX],cnt;

int capital;

inline void Add(int x,int y)
{
	next[++total] = head[x];
	aim[total] = y;
	head[x] = total;
}

void PreDFS(int x,int last)
{
	deep[x] = deep[last] + 1;
	father[x] = last;
	size[x] = 1;
	for(int i = head[x]; i; i = next[i]) {
		if(aim[i] == last)	continue;
		PreDFS(aim[i],x);
		size[x] += size[aim[i]];
		if(size[aim[i]] > size[son[x]])
			son[x] = aim[i];
	}
}

void DFS(int x,int last,int _top)
{
	pos[x] = ++cnt;
	top[x] = _top;
	if(son[x])	DFS(son[x],x,_top);
	for(int i = head[x]; i; i = next[i]) {
		if(aim[i] == last || aim[i] == son[x])	continue;
		DFS(aim[i],x,aim[i]);
	}
}

inline bool InSonTree(int x,int f)
{
	if(pos[x] < pos[f] || pos[x] > pos[f] + size[f] - 1)	return false;
	return true;
}

inline void PushDown(int pos)
{
	if(tree[pos].v) {
		tree[LEFT]._min = tree[pos]._min;
		tree[RIGHT]._min = tree[pos]._min;
		tree[LEFT].v = tree[RIGHT].v = true;
		tree[pos].v = false;
	}
}

void Modify(int l,int r,int x,int y,int pos,int c)
{
	if(l == x && y == r) {
		tree[pos]._min = c;
		tree[pos].v = true;
		return ;
	}
	PushDown(pos);
	int mid = (l + r) >> 1;
	if(y <= mid)	Modify(l,mid,x,y,LEFT,c);
	else if(x > mid)	Modify(mid + 1,r,x,y,RIGHT,c);
	else {
		Modify(l,mid,x,mid,LEFT,c);
		Modify(mid + 1,r,mid + 1,y,RIGHT,c);
	}
	tree[pos]._min = min(tree[LEFT]._min,tree[RIGHT]._min);
}

inline void Modify(int x,int y,int c)
{
	while(top[x] != top[y]) {
		if(deep[top[x]] < deep[top[y]])
			swap(x,y);
		Modify(1,cnt,pos[top[x]],pos[x],1,c);
		x = father[top[x]];
	}
	if(deep[x] < deep[y])	swap(x,y);
	Modify(1,cnt,pos[y],pos[x],1,c);
}

int Ask(int l,int r,int x,int y,int pos)
{
	if(l == x && y == r)	return tree[pos]._min;
	PushDown(pos);
	int mid = (l + r) >> 1;
	if(y <= mid)	return Ask(l,mid,x,y,LEFT);
	if(x > mid)		return Ask(mid + 1,r,x,y,RIGHT);
	int left = Ask(l,mid,x,mid,LEFT);
	int right = Ask(mid + 1,r,mid + 1,y,RIGHT);
	return min(left,right);
}

int main()
{
	cin >> points >> asks;
	for(int x,y,i = 1; i < points; ++i) {
		scanf("%d%d",&x,&y);
		Add(x,y),Add(y,x);
	}
	PreDFS(1,0);
	DFS(1,0,1);
	for(int x,i = 1; i <= cnt; ++i) {
		scanf("%d",&x);
		Modify(1,cnt,pos[i],pos[i],1,x);
	}
	cin >> capital;
	for(int flag,i = 1; i <= asks; ++i) {
		scanf("%d",&flag);
		int x,y,z;
		if(flag == 1) 	scanf("%d",&capital);
		if(flag == 2) {
			scanf("%d%d%d",&x,&y,&z);
			Modify(x,y,z);
		}
		if(flag == 3) {
			scanf("%d",&x);
			if(x == capital)
				printf("%d\n",tree[1]._min);
			else {
				int son = 0;
				for(int j = head[x]; j && !son; j = next[j]) {
					if(aim[j] == father[x])	continue;
					if(InSonTree(capital,aim[j]))
						son = aim[j];
				}
				if(son) {
					int left = Ask(1,cnt,1,pos[son] - 1,1);
					int right = INF;
					if(pos[son] + size[son] <= cnt)
						right = Ask(1,cnt,pos[son] + size[son],cnt,1);
					printf("%d\n",min(min(left,right),Ask(1,cnt,pos[x],pos[x],1)));
				}
				else	printf("%d\n",Ask(1,cnt,pos[x],pos[x] + size[x] - 1,1));
			}
		}
	}
	return 0;
}

BZOJ 3083 遥远的国度 树链剖分

标签：bzoj   树链剖分   子树查询   

原文地址：http://blog.csdn.net/jiangyuze831/article/details/40620585