POJ——T 3728 The merchant

http://poj.org/problem?id=3728

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 5068		Accepted: 1744

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

倍增维护4个值：到当前点的最大钱数，最小钱数，从深度小的点到深度大的点的最大利润，从深度大的点到深度小的点最大利润

统一答案时，分别从u到LCA（u，v），LCA（u，v）到v中找到最优解

（做过原题还WA好多次、、蒟蒻就是蒟蒻，）

  1 #include <cstdio>
  2 
  3 using namespace std;
  4 
  5 const int INF(999999999);
  6 const int N(200000+5);
  7 int pri[N],sumedge,head[N];
  8 struct Edge
  9 {
 10     int v,next;
 11     Edge(int v=0,int next=0):v(v),next(next){}
 12 }edge[N<<1];
 13 inline void ins(int u,int v)
 14 {
 15     edge[++sumedge]=Edge(v,head[u]);
 16     head[u]=sumedge;
 17 }
 18 
 19 #define swap(a,b) {int tmp=a;a=b;b=tmp;}
 20 #define max(a,b) (a>b?a:b)
 21 #define min(a,b) (a<b?a:b)
 22 struct Type_Node
 23 {
 24     int up_down,down_up;
 25     int maxx,minn,dad;
 26 }shop[N][23];
 27 int deep[N];
 28 void DFS(int x,int fa)
 29 {
 30     deep[x]=deep[fa]+1;
 31     shop[x][0].maxx=max(pri[x],pri[fa]);
 32     shop[x][0].minn=min(pri[x],pri[fa]);
 33     shop[x][0].down_up=max(0,pri[fa]-pri[x]);
 34     shop[x][0].up_down=max(0,pri[x]-pri[fa]);
 35     for(int i=1;shop[x][i-1].dad;i++)
 36     {
 37         shop[x][i].dad=shop[shop[x][i-1].dad][i-1].dad;
 38         shop[x][i].maxx=max(shop[x][i-1].maxx,shop[shop[x][i-1].dad][i-1].maxx);
 39         shop[x][i].minn=min(shop[x][i-1].minn,shop[shop[x][i-1].dad][i-1].minn);
 40         shop[x][i].down_up=max(shop[x][i-1].down_up,shop[shop[x][i-1].dad][i-1].down_up);
 41         shop[x][i].down_up=max(shop[x][i].down_up,shop[shop[x][i-1].dad][i-1].maxx-shop[x][i-1].minn);
 42         shop[x][i].up_down=max(shop[x][i-1].up_down,shop[shop[x][i-1].dad][i-1].up_down);
 43         shop[x][i].up_down=max(shop[x][i].up_down,shop[x][i-1].maxx-shop[shop[x][i-1].dad][i-1].minn);
 44     }
 45     for(int i=head[x];i;i=edge[i].next)
 46     {
 47         int v=edge[i].v;
 48         if(shop[x][0].dad!=v) shop[v][0].dad=x,DFS(v,x);
 49     }
 50 }
 51 int LCA(int x,int y)
 52 {
 53     if(deep[x]>deep[y]) swap(x,y);
 54     for(int i=20;i>=0;i--)
 55         if(deep[shop[y][i].dad]>=deep[x]) y=shop[y][i].dad;
 56     if(x==y) return x;
 57     for(int i=20;i>=0;i--)
 58         if(shop[x][i].dad!=shop[y][i].dad) x=shop[x][i].dad,y=shop[y][i].dad;
 59     return shop[x][0].dad; 
 60 }
 61 int Query(int x,int y)
 62 {
 63     int ans=0,maxx=-INF,minn=INF,lca=LCA(x,y);
 64     for(int i=20;i>=0;i--)
 65         if(deep[shop[x][i].dad]>=deep[lca])
 66         {
 67             ans=max(ans,max(shop[x][i].down_up,shop[x][i].maxx-minn));
 68             minn=min(minn,shop[x][i].minn);
 69             x=shop[x][i].dad;
 70         }
 71     for(int i=20;i>=0;i--)
 72         if(deep[shop[y][i].dad]>=deep[lca])
 73         {
 74             ans=max(ans,max(shop[y][i].up_down,maxx-shop[y][i].minn));
 75             maxx=max(maxx,shop[y][i].maxx);
 76             y=shop[y][i].dad;
 77         }
 78     return max(ans,maxx-minn);
 79 }
 80 
 81 inline void read(int &x)
 82 {
 83     x=0;register char ch=getchar();
 84     for(;ch<'0'||ch>'9';) ch=getchar();
 85     for(;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
 86 }
 87 
 88 int main()
 89 {
 90     int n; read(n);
 91     for(int i=1;i<=n;i++) read(pri[i]);
 92     for(int u,v,i=1;i<n;i++)
 93     {
 94         read(u),read(v);
 95         ins(u,v),ins(v,u);
 96     }
 97     DFS(1,0);
 98     int q; read(q);
 99     for(int u,v;q--;)
100     {
101         read(u),read(v);
102         if(u==v) puts("0");
103         else printf("%d
",Query(u,v));
104     }
105     return 0;
106 }

——每当你想要放弃的时候，就想想是为了什么才一路坚持到现在。