POJ 3728

The merchant

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 2887		Accepted: 958

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

lca （trajan)

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 #include <vector>
  6 #include <stack>
  7 
  8 using namespace std;
  9 
 10 const int MAX_N = 50005;
 11 
 12 int N;
 13 int pri[MAX_N],first[MAX_N],v[MAX_N * 2],Next[MAX_N * 2];
 14 int qu[MAX_N],qv[MAX_N];
 15 int Min[MAX_N],Max[MAX_N],down[MAX_N],up[MAX_N];
 16 int f[MAX_N],ans[MAX_N];
 17 bool vis[MAX_N];
 18 vector<int> s[MAX_N];
 19 vector<int> Q[MAX_N];
 20 
 21 void add_edge(int id,int u) {
 22         int e = first[u];
 23         Next[id] = e;
 24         first[u] = id;
 25 }
 26 
 27 int _find(int x) {
 28         if(f[x] == x) return x;
 29 
 30         int t = _find(f[x]);
 31         up[x] = max(up[x],Max[ f[x] ] - Min[ x ]);
 32         up[x] = max(up[x],up[ f[x] ]);
 33         down[x] = max(down[x],down[ f[x] ]);
 34         down[x] = max(down[x],Max[ x ] - Min[ f[x] ]);
 35         Max[x] = max(Max[x],Max[ f[x] ]);
 36         Min[x] = min(Min[x],Min[ f[x] ]);
 37 
 38         return f[x] = t;
 39 }
 40 
 41 void dfs(int u,int fa) {
 42         for(int e = first[u]; e != -1; e = Next[e]) {
 43                 if(v[e] == fa) continue;
 44                 dfs(v[e],u);
 45                 f[_find( v[e] )] = u;
 46         }
 47 
 48         vis[u] = 1;
 49         for(int i = 0; i < Q[u].size(); ++i) {
 50                 int V = qu[ Q[u][i] ] != u ? qu[ Q[u][i] ] : qv[ Q[u][i] ];
 51                 if(vis[V]) {
 52                         int p = _find(V);
 53                         s[p].push_back(Q[u][i]);
 54                 }
 55         }
 56 
 57         for(int i = 0; i < s[u].size(); ++i) {
 58                 int a,b;
 59                 int e = s[u][i];
 60                 a = qu[e],b = qv[e];
 61                 _find(a);
 62                 _find(b);
 63                 ans[e] = max(up[a],down[b]);
 64                 ans[e] = max(ans[e],Max[b] - Min[a]);
 65         }
 66 
 67 
 68 
 69 }
 70 
 71 int main()
 72 {
 73    // freopen("sw.in","r",stdin);
 74     scanf("%d",&N);
 75     for(int i = 1; i <= N; ++i) first[i] = -1;
 76     for(int i = 1; i <= N; ++i) f[i] = i;
 77     for(int i = 1; i <= N; ++i) {
 78             scanf("%d",&pri[i]);
 79             Max[i] = Min[i] = pri[i];
 80     }
 81 
 82     for(int i = 1; i <= 2 * (N - 1); i += 2) {
 83             int u;
 84             scanf("%d%d",&u,&v[i]);
 85             v[i + 1] = u;
 86             add_edge(i,u);
 87             add_edge(i + 1,v[i]);
 88     }
 89 
 90     int q;
 91     scanf("%d",&q);
 92     for(int i = 1; i <= q; ++i) {
 93             scanf("%d%d",&qu[i],&qv[i]);
 94             Q[ qu[i] ].push_back(i);
 95             Q[ qv[i] ].push_back(i);
 96     }
 97 
 98     dfs(1,-1);
 99 
100     for(int i = 1; i <= q; ++i) {
101             printf("%d
",ans[i]);
102     }
103 
104     return 0;
105 }

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