LCA 练习题：【POJ3728】The merchant

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本文介绍了一种使用树形DP算法解决路径最大收益问题的方法。通过预先处理树的结构，利用RMQ技术和LCA算法，快速计算任意两点间路径上的最大盈利。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

const int MAXN = 500010,  INF = 233333333;

struct Edge
{
	int v, id, next;
	Edge(){}
	Edge(int V, int ID, int NEXT) : v(V), id(ID), next(NEXT){}
};

int n, m, size, top;
int uu[MAXN], vv[MAXN], ww[MAXN], anc[MAXN];
int up[MAXN][20], down[MAXN][20], maxw[MAXN][20], minw[MAXN][20], deep[MAXN];
int head[MAXN], head2[MAXN], bin[MAXN], stack[MAXN], mp[MAXN][20], father[MAXN];
bool mark[MAXN];

Edge edge[MAXN * 2], edge2[MAXN * 2];

void Init(int num)
{
	for (int i = 0; i <= num; ++ i) head[i] = head2[i] =  - 1, mark[i] = false;
	size = top = 0;
}

void Insert(int u, int v, int id)
{
	edge[size] = Edge(v, id, head[u]);
	head[u] = size ++ ;
}

void Insert2(int u, int v, int id)
{
	edge2[size] = Edge(v, id, head2[u]);
	head2[u] = size ++ ;
}

void dfs(int u, int father, int k)
{
	deep[u] = k;
	for (int i = head[u]; i != - 1; i = edge[i].next){
		int v = edge[i].v;
		if (v == father) continue;
		dfs(v, u, k + 1);
	}
}

void RMQ(int u, int father)
{
	stack[ ++ top] = u;
	int fa = stack[top - 1];
	up[u][0] = down[u][0] = 0;
	maxw[u][0] = minw[u][0] = ww[u];
	for (int i = 1;bin[i] <= top; ++ i)
    {
		fa = stack[top - bin[i - 1]];
		up[u][i] = max(max(up[u][i - 1], up[fa][i - 1]), maxw[fa][i - 1] - minw[u][i - 1]);
		down[u][i] = max(max(down[u][i - 1], down[fa][i - 1]), maxw[u][i - 1] - minw[fa][i - 1]);
		maxw[u][i] = max(maxw[u][i - 1], maxw[fa][i - 1]);
		minw[u][i] = min(minw[u][i - 1], minw[fa][i - 1]);
		mp[u][i] = stack[top - bin[i]];
	}
	for (int i = head[u]; i != -1; i = edge[i].next)
	{
		int v = edge[i].v;
		if (v == father) continue;
		RMQ(v, u);
	}
	--top;
}

int findset(int v)
{
	if (v !=  father[v]) father[v] = findset(father[v]);
	return father[v];
}

void LCA(int u)
{
	mark[u] = true;
	father[u] = u;
	for (int i = head2[u]; i != -1; i = edge2[i].next)
    {
		int v = edge2[i].v, id = edge2[i].id;
		if (!mark[v]) continue;
		anc[id] = findset(v);
	}
	for (int i = head[u]; i != -1; i = edge[i].next)
	{
		int v = edge[i].v;
		if (mark[v]) continue;
		LCA(v);
		father[v] = u;
	}
}

int search(int x)
{
	int i = 0;
	while (bin[i + 1] <= x) ++ i;
	return i;
}

int Minw(int u, int anc)
{
	int i = search(deep[u] - deep[anc] + 1);
	if (bin[i] == deep[u] - deep[anc] + 1) return minw[u][i];
	return min(minw[u][i], Minw(mp[u][i], anc));
}

int Maxw(int u, int anc)
{
	int i = search(deep[u] - deep[anc] + 1);
	if (bin[i] == deep[u] - deep[anc] + 1) return maxw[u][i];
	return max(maxw[u][i], Maxw(mp[u][i], anc));
}

int PushUp(int u, int anc)
{
	int i = search(deep[u] - deep[anc] + 1);
	if (bin[i] == deep[u] - deep[anc] + 1) return up[u][i];
	int upfa = PushUp(mp[u][i], anc);
	upfa = max(upfa, up[u][i]);
	int maxwfa = Maxw(mp[u][i], anc);
	return max(upfa, maxwfa - minw[u][i]);
}

int PushDown(int u, int anc)
{
	int i = search(deep[u] - deep[anc] + 1);
	if (bin[i] == deep[u] - deep[anc] + 1) return down[u][i];
	int downfa = PushDown(mp[u][i], anc);
	downfa = max(downfa, down[u][i]);
	int minwfa = Minw(mp[u][i], anc);
	return max(downfa, maxw[u][i] - minwfa);
}

int main(void)
{
	bin[0] = 1;
	for (int i = 1; bin[i - 1] < MAXN; ++i)bin[i] = bin[i - 1] * 2;
	int u, v;
	while (scanf("%d", &n) == 1)
    {
		Init(n);
		for (int i = 1; i <= n; ++ i) scanf("%d", ww + i);
		for (int i = 1; i < n; ++ i)
        {
			scanf("%d%d", &u, &v);
			Insert(u, v, i);
			Insert(v, u, i);
		}
		size = 0;
		scanf("%d", &m);
		for (int i = 0; i < m; ++ i)
		{
			scanf("%d%d", uu + i, vv + i);
			Insert2(uu[i], vv[i], i);
			Insert2(vv[i], uu[i], i);
		}
		dfs(1, -1, 1);
		RMQ(1, -1);
		LCA(1);
		for (int i = 0; i < m; ++ i)
		{
			int upmax = PushUp(uu[i], anc[i]), downmax = PushDown(vv[i], anc[i]);
			int Minww = Minw(uu[i], anc[i]), Maxww = Maxw(vv[i], anc[i]);
			printf("%d\n", max(max(upmax, downmax), Maxww - Minww));
		}
	}
	return 0;
}

原题： http://poj.org/problem?id=3728

题意：给出一棵节点有值的树，给出Q个询问(a,b)，问从a到b的最大盈利（即：先在最小值买入，再在最大值卖出）

The merchant

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 3344		Accepted: 1110

Description

There are N cities in a country, and there is one and only one simple path between each pair of cities. A merchant has chosen some paths and wants to earn as much money as possible in each path. When he move along a path, he can choose one city to buy some goods and sell them in a city after it. The goods in all cities are the same but the prices are different. Now your task is to calculate the maximum possible profit on each path.

Input

The first line contains N, the number of cities.
Each of the next N lines contains w_i the goods' price in each city.
Each of the next N-1 lines contains labels of two cities, describing a road between the two cities.
The next line contains Q, the number of paths.
Each of the next Q lines contains labels of two cities, describing a path. The cities are numbered from 1 to N.

1 ≤ N, w_i, Q ≤ 50000

Output

The output contains Q lines, each contains the maximum profit of the corresponding path. If no positive profit can be earned, output 0 instead.

Sample Input

Sample Output

Source

POJ Monthly Contest – 2009.04.05, GaoYihan