POJ

若买卖不在相邻城市，朴素的想法是遍历找出路径，然后找出路径上的最高最低价格得到价格差。然而每次光找路径就要耗费O(n)的时间，而题目中肯定有多个Query。

用动态规划的思路解决问题，对指定两点uv，假设v往上走2^k步的父节点为t，走2^(k+1)步的父节点为u，如下图所示

边上的数字表示深度之差。

定义四个dp数组：

int dp_max[MAX_LOG_V][MAX_V], dp_min[MAX_LOG_V][MAX_V]; // 往上走2^k步之间的最高与最低价格
int dp_up[MAX_LOG_V][MAX_V], dp_down[MAX_LOG_V][MAX_V]; // [从v往上走2^k步之间]或[往下走2^k步到v之间]相应的最大利润
前两个很好计算，对于uv的dp_up，买卖事件有3种情况：

买和卖都发生在ut之间

买和卖都发生在tv之间

买和卖分别发生在ut和tv之间

所以递推公式如下

dp_up[k + 1][v] = max(max(dp_up[k][v], dp_up[k][t]), dp_max[k][t] - dp_min[k][v]);
同理有dp_down。

预处理算出这4个dp数组后，对任意uv，取t=lca，用类似的思想分成3种情况（对于前两种情况有down和up的子分支），取最大值即可。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>using namespace std;const int MaxN = 50000;const int MaxLog = 16;int n, m;vector<int> g[MaxN + 1];int w[MaxN + 1], f[MaxLog][MaxN + 1], depth[MaxN + 1];int dp_max[MaxLog][MaxN + 1], dp_min[MaxLog][MaxN + 1];int dp_up[MaxLog][MaxN + 1], dp_down[MaxLog][MaxN + 1];void dfs(int u, int fa, int deep){    f[0][u] = fa;    depth[u] = deep;    dp_max[0][u] = max(w[u], w[fa]);    dp_min[0][u] = min(w[u], w[fa]);    dp_up[0][u] = max(w[fa] - w[u], 0);    dp_down[0][u] = max(w[u] - w[fa], 0);    for (int i = 0; i < g[u].size(); i++)    {        int v = g[u][i];        if (v != fa)            dfs(v, u, deep + 1);    }}void init(){    memset(dp_max, 0, sizeof(dp_max));    memset(dp_min, 0x3f, sizeof(dp_min));    dfs(1, 0, 1);    for (int k = 0; k + 1 < MaxLog; k++)        for (int v = 1; v <= n; v++)        {            if (!f[k][v])                f[k + 1][v] = 0;            else            {                f[k + 1][v] = f[k][f[k][v]];                int t = f[k][v];                dp_max[k + 1][v] = max(dp_max[k][v], dp_max[k][t]);                dp_min[k + 1][v] = min(dp_min[k][v], dp_min[k][t]);                dp_up[k + 1][v] = max(max(dp_up[k][v], dp_up[k][t]), dp_max[k][t] - dp_min[k][v]);                dp_down[k + 1][v] = max(max(dp_down[k][v], dp_down[k][t]), dp_max[k][v] - dp_min[k][t]);            }        }}int lca(int a, int b){    if (depth[a] < depth[b])        swap(a, b);    int d = depth[a] - depth[b];    for (int i = 0; d; d >>= 1, i++)        if (d & 1)            a = f[i][a];    if (a == b)        return a;    for (int i = MaxLog - 1; i >= 0; i--)        if (f[i][a] != f[i][b])            a = f[i][a], b = f[i][b];    return f[0][a];}int up(int x, int k, int &min_price){    min_price = 0x3f3f3f3f;    int max_profit = 0;    int prev_min_price = 0x3f3f3f3f;    for (int i = MaxLog - 1; i >= 0; i--)    {        if (k >> i & 1)        {            min_price = min(min_price, dp_min[i][x]);            max_profit = max(max_profit, dp_up[i][x]);            max_profit = max(max_profit, dp_max[i][x] - prev_min_price);            prev_min_price = min(prev_min_price, dp_min[i][x]);            x = f[i][x];        }    }    return max_profit;}int down(int x, int k, int &max_price){    max_price = 0;    int max_profit = 0;    int pre_max_price = 0;    for (int i = MaxLog - 1; i >= 0; i--)    {        if (k >> i & 1)        {            max_price = max(max_price, dp_max[i][x]);            max_profit = max(max_profit, dp_down[i][x]);            max_profit = max(max_profit, pre_max_price - dp_min[i][x]);            pre_max_price = max(pre_max_price, dp_max[i][x]);            x = f[i][x];        }    }    return max_profit;}int main(){    scanf("%d", &n);    for (int i = 1; i <= n; i++)        scanf("%d", &w[i]);    for (int i = 1; i <= n; i++)        g[i].clear();    for (int i = 1; i < n; i++)    {        int u, v;        scanf("%d %d", &u, &v);        g[u].push_back(v);        g[v].push_back(u);    }    init();    int Q;    scanf("%d", &Q);    while (Q--)    {        int u, v;        scanf("%d %d", &u, &v);        int k = lca(u, v);        int max_price, min_price;        int up_profit = up(u, depth[u] - depth[k], min_price);        int down_profit = down(v, depth[v] - depth[k], max_price);        int ans = max(max(up_profit, down_profit), max_price - min_price);        printf("%d\n", ans);    }    return 0;}