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 (ui,vi) for the value of f(ui,vi). Help bobo answer them.
Let f(a,b) be the minimum of vertices not on the path between vertices a and b.
There are q queries (ui,vi) for the value of f(ui,vi). Help bobo answer them.
Input
The input consists of several tests. For each tests:
The first line contains 2 integers n,q (4≤n≤106,1≤q≤106). Each of the following (n - 1) lines contain 2 integers ai,bi denoting an edge between vertices ai and bi (1≤ai,bi≤n). Each of the following q lines contains 2 integer u′i,v′i (1≤ui,vi≤n).
The queries are encrypted in the following manner.
u1=u′1,v1=v′1.
For i≥2, ui=u′i⊕f(ui - 1,vi - 1),vi=v′i⊕f(ui-1,vi-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.
The first line contains 2 integers n,q (4≤n≤106,1≤q≤106). Each of the following (n - 1) lines contain 2 integers ai,bi denoting an edge between vertices ai and bi (1≤ai,bi≤n). Each of the following q lines contains 2 integer u′i,v′i (1≤ui,vi≤n).
The queries are encrypted in the following manner.
u1=u′1,v1=v′1.
For i≥2, ui=u′i⊕f(ui - 1,vi - 1),vi=v′i⊕f(ui-1,vi-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.
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，然后增加不经过1，那么答案直接为1，否则就是预处理，
数据规模非常大，所以仅仅能用bfs，而且须要加读写外挂，具体过程代码具体解释。
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;//假设不经过1，而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++;//把包括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
",last);
}
}
return 0;
}