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.
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.
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\n",last); } } return 0; }