POJ 3278：The merchant（LCA&DP）

The merchant

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 6864		Accepted: 2375

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

题意

给出一棵树，每个节点有一个点权，代表商品的价值

从节点u->v的路径上，有一个商人从一个节点购买商品，并在另一个节点卖出，问商人可以获得的最大利润是多少

思路

商人进行买卖一共有三种情况：

在点u->lca(u,v)的路上买，lca(u,v)->v的路上卖
在u->lca(u,v)的路上完成买卖
在lca(u,v)->v的路上完成买卖

所以我们可以引入四个数组：up，down，maxx，minn。分别代表：在u->lca(u,v)的路上完成买卖的最大获利；在lca(u,v)->v的路上完成买卖的最大获利；lca(u,v)->v点出售的最大利润；u->lca(u,v)购买的最少花费

我们可以得到商人的最大收益=max(max(up[u],down[v]),maxx[v]-minn[u])

问题就变成了去寻找lca(u,v)，并更新数组的值。以上四个数组的值可以在并查集find函数中进行更新

代码

  1 #include <iostream>
  2 #include <algorithm>
  3 #include <cstring>
  4 #define ll long long
  5 #define ull unsigned long long
  6 #define ms(a,b) memset(a,b,sizeof(a))
  7 const int inf=0x3f3f3f3f;
  8 const ll INF=0x3f3f3f3f3f3f3f3f;
  9 const int maxn=1e6+10;
 10 const int mod=1e9+7;
 11 const int maxm=1e3+10;
 12 using namespace std;
 13 int maxx[maxn],minn[maxn];
 14 int up[maxn],down[maxn];
 15 int f[maxn];
 16 // minn[u]        u->lca(u,v)买的最小值
 17 // maxx[v]        lca(u,v)->v卖的最大值
 18 // up[u]           u->lca(u,v)进行买卖的最大值
 19 // down[v]       lca(u,v)->v进行买卖的最大值
 20 int find(int x)
 21 {
 22     if(f[x]==x)
 23         return x;
 24     int y=f[x];
 25     f[x]=find(f[x]);
 26     up[x]=max(max(up[y],up[x]),maxx[y]-minn[x]);
 27     down[x]=max(max(down[y],down[x]),maxx[x]-minn[y]);
 28     maxx[x]=max(maxx[x],maxx[y]);
 29     minn[x]=min(minn[x],minn[y]);
 30     return f[x];
 31 }
 32 void join(int x,int y)
 33 {
 34     int dx=f[x],dy=f[y];
 35     if(dx!=dy)
 36         f[y]=dx;
 37 }
 38 struct Edge
 39 {
 40     int to,Next;
 41 }edge[maxn];
 42 int head1[maxn];
 43 int tot1;
 44 void add_edge(int u,int v)
 45 {
 46     edge[tot1].to=v;
 47     edge[tot1].Next=head1[u];
 48     head1[u]=tot1++;
 49 }
 50 struct Query
 51 {
 52     int to,Next;
 53     int index;
 54 }query[maxn];
 55 int head2[maxn];
 56 int tot2;
 57 void add_query(int u,int v,int index)
 58 {
 59     query[tot2].to=v;
 60     query[tot2].Next=head2[u];
 61     query[tot2].index=index;
 62     head2[u]=tot2++;
 63 }
 64 struct Pos
 65 {
 66     int to,Next;
 67     int index;
 68 }pos[maxn];
 69 int head3[maxn];
 70 int tot3;
 71 void add_pos(int u,int v,int index)
 72 {
 73     pos[tot3].to=v;
 74     pos[tot3].Next=head3[u];
 75     pos[tot3].index=index;
 76     head3[u]=tot3++;
 77 }
 78 int vis[maxn];
 79 int qu[maxn],qv[maxn];
 80 int fa[maxn];
 81 int ans[maxn];
 82 void LCA(int u)
 83 {
 84     fa[u]=u;
 85     vis[u]=1;
 86     for(int i=head1[u];~i;i=edge[i].Next)
 87     {
 88         int v=edge[i].to;
 89         if(vis[v])
 90             continue;
 91         LCA(v);
 92         join(u,v);
 93         fa[find(u)]=u;
 94     }
 95     for(int i=head2[u];~i;i=query[i].Next)
 96     {
 97         int v=query[i].to;
 98         if(vis[v])
 99         {
100             int t=fa[find(v)];
101             // 储存lca(u,v)的子节点
102             add_pos(t,v,query[i].index);
103         }
104     }
105     for(int i=head3[u];~i;i=pos[i].Next)
106     {
107         int id=pos[i].index;
108         int xx=qu[id],yy=qv[id];
109         // 更新四个数组的值
110         find(xx);find(yy);
111         ans[id]=max(max(up[xx],down[yy]),maxx[yy]-minn[xx]);
112     }
113 }
114 int value[maxn];
115 void init(int n)
116 {
117     tot1=0;tot2=0;tot3=0;
118     for(int i=1;i<=n;i++)
119         f[i]=i;
120     ms(fa,0);ms(vis,0);
121     ms(head1,-1);ms(head2,-1);ms(head3,-1);
122 }
123 int main(int argc, char const *argv[])
124 {
125     #ifndef ONLINE_JUDGE
126         freopen("/home/wzy/in.txt", "r", stdin);
127         freopen("/home/wzy/out.txt", "w", stdout);
128         srand((unsigned int)time(NULL));
129     #endif
130     ios::sync_with_stdio(false);
131     cin.tie(0);
132     int n;
133     cin>>n;
134     init(n);
135     for(int i=1;i<=n;i++)
136         cin>>value[i],minn[i]=value[i],maxx[i]=value[i];
137     int u,v;
138     for(int i=1;i<n;i++)
139         cin>>u>>v,add_edge(u,v),add_edge(v,u);
140     int q;
141     cin>>q;
142     for(int i=0;i<q;i++)
143         cin>>qu[i]>>qv[i],add_query(qu[i],qv[i],i),add_query(qv[i],qu[i],i);
144     LCA(1);
145     for(int i=0;i<q;i++)
146         cout<<ans[i]<<"
";
147     #ifndef ONLINE_JUDGE
148         cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
149     #endif
150     return 0;
151 }