HDU 4916 树形dp

Count on the path

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 92    Accepted Submission(s): 10

Problem Description

bobo has a tree, whose vertices are conveniently labeled by 1,2,…,n.

Let f(a,b) be the minimum of vertices not on the path between vertices a and b.

There are q queries (u _i,v _i) for the value of f(u _i,v _i). Help bobo answer them.

 

Input

The input consists of several tests. For each tests:

The first line contains 2 integers n,q (4≤n≤10 ⁶,1≤q≤10 ⁶). Each of the following (n - 1) lines contain 2 integers a _i,b _i denoting an edge between vertices a _i and b _i (1≤a _i,b _i≤n). Each of the following q lines contains 2 integer u′ _i,v′ _i (1≤u _i,v _i≤n).

The queries are encrypted in the following manner.

u ₁=u′ ₁,v ₁=v′ ₁.
For i≥2, u _i=u′ _i⊕f(u _{i - 1},v _{i - 1}),v _i=v′ _i⊕f(u _i-1,v _i-1).

Note ⊕ denotes bitwise exclusive-or.

It is guaranteed that f(a,b) is defined for all a,b.

The task contains huge inputs. `scanf` in g++ is considered too slow to get accepted. You may (1) submit the solution in c++; or (2) use hand-written input utilities.

 

Output

For each tests:

For each queries, a single number denotes the value.

 

Sample Input

  
  

   
   4 1
1 2
1 3
1 4
2 3
5 2
1 2
1 3
2 4
2 5
1 2
7 6
  
  

 

Sample Output

  
  

   
   4
3
1
  
  

 

Author

Xiaoxu Guo (ftiasch)

给定一棵树，求不经过路径的最小标号。

把1作为根，然后加入不经过1，那么答案直接为1，否则就是预处理，

数据规模很大，所以只能用bfs，并且需要加读写外挂，具体过程代码详解。

代码：

/* ***********************************************
Author :rabbit
Created Time :2014/8/6 10:44:17
File Name :5.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
int fun(){
       char ch;int flag=1,a=0;
       while(ch=getchar())if((ch>='0'&&ch<='9')||ch=='-')break;
       if(ch=='-')flag=-1;else a=ch-'0';
       while(ch=getchar()){
              if(ch>='0'&&ch<='9')a=10*a+ch-'0';
              else break;
       }
       return flag*a;
}
const int maxn=1001000;
int head[maxn],tol;
int subtree[maxn];//子树最小标号
int belong[maxn];//所在的与根节点1相连的子树最小标号。
int child[maxn][4];//儿子子树前4小。
int que[maxn];//广搜队列。
int path[maxn];//path[u]代表u所在的在根节点1的儿子的子树中从根节点到u路径以外的最小标号。
int fa[maxn];//父亲标号。
int dep[maxn];//深度数组
struct Edge{
	int next,to;
}edge[2*maxn];
void addedge(int u,int v){
	edge[tol].to=v;
	edge[tol].next=head[u];
	head[u]=tol++;
}
int main()
{
     //freopen("data.in","r",stdin);
     //freopen("data.out","w",stdout);
     int n,m;
	 while(~scanf("%d%d",&n,&m)){
		 memset(head,-1,sizeof(head));tol=0;
		 for(int i=1;i<n;i++){
			 int u,v;
			 u=fun();v=fun();
			 addedge(u,v);
			 addedge(v,u);
		 }
		 int front=0,rear=0;
		 dep[1]=0;fa[1]=-1;
		 que[rear++]=1;
		 while(front!=rear){
			 int u=que[front++];
			 for(int i=head[u];i!=-1;i=edge[i].next){
				 int v=edge[i].to;
				 if(v==fa[u])continue;
				 dep[v]=dep[u]+1;
				 fa[v]=u;
				 que[rear++]=v;
			 }
		 }
		 for(int i=1;i<=n;i++)
			 for(int j=0;j<4;j++)
				 child[i][j]=INF;
		 for(int i=rear-1;i>=0;i--){
			 int u=que[i];
			 subtree[u]=min(u,child[u][0]);
			 int p=fa[u];
			 if(p==-1)continue;
			 child[p][3]=subtree[u];
			 sort(child[p],child[p]+4);
		 }
		 front=0,rear=0;
		 path[1]=INF;
		 belong[1]=-1;
		 for(int i=head[1];i!=-1;i=edge[i].next){
			 int u=edge[i].to;
			 path[u]=INF;
			 belong[u]=subtree[u];
			 que[rear++]=u;
		 }
		 while(front!=rear){
			 int u=que[front++];
			 for(int i=head[u];i!=-1;i=edge[i].next){
				 int v=edge[i].to;
				 if(v==fa[u])continue;
				 path[v]=min(path[u],child[u][subtree[v]==child[u][0]]);
				 belong[v]=belong[u];
				 que[rear++]=v;
			 }
			 path[u]=min(path[u],child[u][0]);//u的儿子子树的最小标号。
		 }
		 int last=0;
		 while(m--){
			 int u,v;
			 u=fun();v=fun();
			 u^=last;v^=last;
			 if(u>v)swap(u,v);
			 if(u!=1&&belong[u]==belong[v])last=1;//如果不经过1，并且属于同一个根节点的儿子的子树，答案直接为1
			 else{
				 int i=0;
				 while(child[1][i]==belong[u]||child[1][i]==belong[v])i++;//把包含u，v的儿子子树跳过。
				 last=u==1?path[v]:min(path[u],path[v]);//路径分为两段，取最小值。
				 last=min(last,child[1][i]);//出去u，v所在的子树以外的最小值。
			 }
			 printf("%d\n",last);
		 }
	 }
     return 0;
}