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≤10⁶,1≤q≤10⁶). 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.

u₁=u′₁,v₁=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 11 21 31 42 35 21 21 32 42 51 27 6

 

Sample Output

431

 

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 :rabbitCreated Time :2014/8/6 10:44:17File 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;}