hdu 4916 Count on the path
2014-08-08 22:26
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Count on the path
Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 378 Accepted Submission(s): 113
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)
Source
2014 Multi-University Training Contest 5
题意:给你一棵树,最多有1e6个点。不断询问a->b外的最小点。
基本仿照gxx的标程来写的,感觉学到了很多。这题主要是考一些树上的细腻的操作。
思路见注释
代码:
#include <iostream> #include <algorithm> #include <cstdio> #include <string> #include <cstring> #include <cmath> #include <vector> #include <list> #include <map> #include <set> #include <deque> #include <queue> #include <stack> #include <bitset> #include <functional> #include <sstream> #include <iomanip> #include <cmath> #include <cstdlib> #include <ctime> #pragma comment(linker, "/STACK:102400000,102400000") typedef long long ll; #define INF 1e9 #define MAXN 21 const int maxn = 1000005; #define mod 1000000007 #define eps 1e-7 #define pi 3.1415926535897932384626433 #define rep(i,n) for(int i=0;i<n;i++) #define rep1(i,n) for(int i=1;i<=n;i++) #define scan(n) scanf("%d",&n) #define scanll(n) scanf("%I64d",&n) #define scan2(n,m) scanf("%d%d",&n,&m) #define scans(s) scanf("%s",s); #define ini(a) memset(a,0,sizeof(a)) #define out(n) printf("%d\n",n) //ll gcd(ll a,ll b) { return b==0?a:gcd(b,a%b);} using namespace std; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 int q[maxn]; int path[maxn]; //path[i]记录表示在i所在的子树中(以1为根),除i到1这条路径外的最小值 int child[maxn][4]; //child[i][0]表示以i为根且不包括i的树的最小点,child[i][1]为次小,child[i][2]为第三小 int fa[maxn]; int belong[maxn]; //belong[i]表示i所在的子树(以1为根)的最小值,可用来快速判断两点是否位于同一子树 int subtree[maxn];//subtree[i]表示以i为根(包括i)的子树的最小值 vector<int> G[maxn]; //邻接表 int n,m; int next_int() { int result = 0; char c = getchar(); while (!isdigit(c)) { c = getchar(); } while (isdigit(c)) { result = result * 10 + c - '0'; c = getchar(); } return result; } int query(int a,int b) { if(a > b) swap(a,b); if(a != 1 && (belong[a] == belong[b])) return 1; //若两点位于同一子树,则不经过1 int i = 0; while(child[1][i] == belong[a] || child[1][i] == belong[b]) i++; //找出除a,b所在子树外最小的点 int ans = a == 1 ? path[b] : min(path[a], path[b]); //a,b所在子树最小的点(a->1,b->1的路径除外) ans = min(ans, child[1][i]); return ans; } int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); // freopen("out.txt","w",stdout); #endif while(~scanf("%d%d",&n,&m)) { int a,b; rep(i,n+1) G[i].clear(); rep(i,n-1) { a = next_int(); b = next_int(); G[a].push_back(b); //用了vector保存邻接表,不断pb g++T了,用C++过了 G[b].push_back(a); } int head = 0, tail = 0; q[tail++] = 1; fa[1] = -1; while(head < tail) { int u = q[head++]; for(int i = 0;i <(int)G[u].size(); i++) { int v = G[u][i]; if(v == fa[u]) continue; fa[v] = u; q[tail++] = v; } } rep1(i,n) rep(j,4) child[i][j] = INF; for(int i = tail-1; i >= 0; i--) { int v = q[i]; subtree[v] = min(child[v][0], v); int u = fa[v]; if(u != -1) { child[u][3] = subtree[v]; sort(child[u], child[u] + 4); //path[v] = min(child[u][0] } } head = tail = 0; for(int i = 0;i < (int)G[1].size(); i++) { int u = G[1][i]; belong[u] = subtree[u]; path[u] = INF; q[tail++] = u; } path[1] = INF; belong[1] = -1; while(head < tail) { int u = q[head++]; for(int i = 0;i < (int)G[u].size(); i++) { int v = G[u][i]; if(v == fa[u]) continue; belong[v] = belong[u]; path[v] = min(path[u], child[u][subtree[v] == child[u][0]]); q[tail++] = v; } path[u] = min(path[u], child[u][0]); } int last = 0; while(m--) { a = next_int(); b = next_int(); a ^= last; b ^= last; int ans = query(a,b); last = ans; printf("%d\n",ans); } } return 0; }
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