POJ 1986:Distance Queries
Distance Queries
Time Limit: 2000MS | Memory Limit: 30000K | |
Total Submissions: 18139 | Accepted: 6248 | |
Case Time Limit: 1000MS |
Description
Farmer John's cows refused to run in his marathon since he chose a path much too long for their leisurely lifestyle. He therefore wants to find a path of a more reasonable length. The input to this problem consists of the same input as in "Navigation Nightmare",followed by a line containing a single integer K, followed by K "distance queries". Each distance query is a line of input containing two integers, giving the numbers of two farms between which FJ is interested in computing distance (measured in the length of the roads along the path between the two farms). Please answer FJ's distance queries as quickly as possible!
Input
* Lines 1..1+M: Same format as "Navigation Nightmare"
* Line 2+M: A single integer, K. 1 <= K <= 10,000
* Lines 3+M..2+M+K: Each line corresponds to a distance query and contains the indices of two farms.
Output
* Lines 1..K: For each distance query, output on a single line an integer giving the appropriate distance.
Sample Input
7 6
1 6 13 E
6 3 9 E
3 5 7 S
4 1 3 N
2 4 20 W
4 7 2 S
3
1 6
1 4
2 6
Sample Output
13
3
36
Hint
Farms 2 and 6 are 20+3+13=36 apart.
题意
n个点,m条边,每两个相连的点有一个距离,对于每次询问,求出u,v的距离
思路
因为题中给出的图是一个树(Navigation Nightmare题目链接:http://poj.org/problem?id=1984)
对于树上的两点距离,我们有:dis(u,v)=dis(u,root)+dis(v,toot)-2*dis(lca(u,v),root)
预处理出来每个点到根节点的距离,在查询的时候求出u,v两点的lca,然后利用上述公式计算即可
因为是一棵树,所以可以以任意一个节点作为根节点
代码
1 #include <algorithm>
2 #include <iostream>
3 #include <string.h>
4 #define ll long long
5 #define ull unsigned long long
6 #define ms(a,b) memset(a,b,sizeof(a))
7 const int inf=0x3f3f3f3f;
8 const ll INF=0x3f3f3f3f3f3f3f3f;
9 const int maxn=2e5+10;
10 const int mod=1e9+7;
11 const int maxm=1e3+10;
12 using namespace std;
13 int f[maxn];
14 int find(int x)
15 {
16 if(f[x]!=x)
17 f[x]=find(f[x]);
18 return f[x];
19 }
20 inline void join(int x,int y)
21 {
22 int dx=f[x],dy=f[y];
23 if(dx!=dy)
24 f[dy]=dx;
25 }
26 struct Edge
27 {
28 int to,Next;
29 int value;
30 }edge[maxn];
31 int tot1;
32 int head1[maxn];
33 inline void add_edge(int u,int v,int w)
34 {
35 edge[tot1].to=v;
36 edge[tot1].value=w;
37 edge[tot1].Next=head1[u];
38 head1[u]=tot1++;
39 }
40 int vist[maxn];
41 int dis[maxn];
42 // 预处理每个点到根节点的距离
43 void dfs(int u,int len)
44 {
45 dis[u]=len;
46 vist[u]=1;
47 for(int i=head1[u];~i;i=edge[i].Next)
48 {
49 int v=edge[i].to;
50 if(!vist[v])
51 dfs(v,len+edge[i].value);
52 }
53 }
54 struct Query
55 {
56 int to,nex;
57 int index;
58 }query[maxn];
59 int head2[maxn];
60 int tot2;
61 inline void add_query(int u,int v,int index)
62 {
63 query[tot2].to=v;
64 query[tot2].index=index;
65 query[tot2].nex=head2[u];
66 head2[u]=tot2++;
67 query[tot2].to=u;
68 query[tot2].index=index;
69 query[tot2].nex=head2[v];
70 head2[v]=tot2++;
71 }
72 int vis[maxn];
73 int fa[maxn];
74 int ans[maxn];
75 void LCA(int u)
76 {
77 fa[u]=u;
78 vis[u]=1;
79 for(int i=head1[u];~i;i=edge[i].Next)
80 {
81 int v=edge[i].to;
82 if(vis[v])
83 continue;
84 LCA(v);
85 join(u,v);
86 fa[find(u)]=u;
87 }
88 for(int i=head2[u];~i;i=query[i].nex)
89 {
90 int v=query[i].to;
91 if(vis[v])
92 ans[query[i].index]=fa[find(v)];
93 }
94 }
95 inline void init(int n)
96 {
97 tot1=tot2=0;
98 ms(head1,-1);
99 ms(head2,-1);
100 ms(vis,0);
101 ms(vist,0);
102 ms(fa,0);
103 ms(dis,0);
104 for(int i=1;i<=n;i++)
105 f[i]=i;
106 }
107 int x[maxn],y[maxn];
108 int main(int argc, char const *argv[])
109 {
110 #ifndef ONLINE_JUDGE
111 freopen("/home/wzy/in.txt", "r", stdin);
112 freopen("/home/wzy/out.txt", "w", stdout);
113 srand((unsigned int)time(NULL));
114 #endif
115 ios::sync_with_stdio(false);
116 cin.tie(0);
117 int n,m,q;
118 while(cin>>n>>m)
119 {
120 init(n);
121 int u,v,w;
122 char ch[3];
123 for(int i=0;i<m;i++)
124 cin>>u>>v>>w>>ch,add_edge(u,v,w),add_edge(v,u,w);
125 dfs(1,0);
126 cin>>q;
127 for(int i=0;i<q;i++)
128 cin>>x[i]>>y[i],add_query(x[i],y[i],i);
129 LCA(1);
130 for(int i=0;i<q;i++)
131 cout<<dis[x[i]]+dis[y[i]]-2*dis[ans[i]]<<endl;
132 }
133 #ifndef ONLINE_JUDGE
134 cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
135 #endif
136 return 0;
137 }
POJ 1986:Distance Queries的更多相关文章
- POJ 1986:Distance Queries(倍增求LCA)
http://poj.org/problem?id=1986 题意:给出一棵n个点m条边的树,还有q个询问,求树上两点的距离. 思路:这次学了一下倍增算法求LCA.模板. dp[i][j]代表第i个点 ...
- poj-1986 Distance Queries(lca+ST+dfs)
题目链接: Distance Queries Time Limit: 2000MS Memory Limit: 30000K Total Submissions: 11531 Accepted ...
- POJ 1986 Distance Queries(Tarjan离线法求LCA)
Distance Queries Time Limit: 2000MS Memory Limit: 30000K Total Submissions: 12846 Accepted: 4552 ...
- poj 1986 Distance Queries LCA
题目链接:http://poj.org/problem?id=1986 Farmer John's cows refused to run in his marathon since he chose ...
- POJ 1986 - Distance Queries - [LCA模板题][Tarjan-LCA算法]
题目链接:http://poj.org/problem?id=1986 Description Farmer John's cows refused to run in his marathon si ...
- POJ 1986 Distance Queries 【输入YY && LCA(Tarjan离线)】
任意门:http://poj.org/problem?id=1986 Distance Queries Time Limit: 2000MS Memory Limit: 30000K Total ...
- POJ 1986 Distance Queries LCA两点距离树
标题来源:POJ 1986 Distance Queries 意甲冠军:给你一棵树 q第二次查询 每次你问两个点之间的距离 思路:对于2点 u v dis(u,v) = dis(root,u) + d ...
- POJ 1986 Distance Queries / UESTC 256 Distance Queries / CJOJ 1129 【USACO】距离咨询(最近公共祖先)
POJ 1986 Distance Queries / UESTC 256 Distance Queries / CJOJ 1129 [USACO]距离咨询(最近公共祖先) Description F ...
- POJ.1986 Distance Queries ( LCA 倍增 )
POJ.1986 Distance Queries ( LCA 倍增 ) 题意分析 给出一个N个点,M条边的信息(u,v,w),表示树上u-v有一条边,边权为w,接下来有k个询问,每个询问为(a,b) ...
随机推荐
- 06 windows安装Python+Pycharm+Scrapy环境
windows安装Python+Pycharm+Scrapy环境 使用微信扫码关注微信公众号,并回复:"Python工具包",免费获取下载链接! 一.卸载python环境 卸载以下 ...
- 零基础学习java------37---------mybatis的高级映射(单表查询,多表(一对一,一对多)),逆向工程,Spring(IOC,DI,创建对象,AOP)
一. mybatis的高级映射 1 单表,字段不一致 resultType输出映射: 要求查询的字段名(数据库中表格的字段)和对应的java类型的属性名一致,数据可以完成封装映射 如果字段和jav ...
- Use of explicit keyword in C++
Predict the output of following C++ program. 1 #include <iostream> 2 3 using namespace std; 4 ...
- SpringIOC原理浅析
1. IoC理论的背景我们都知道,在采用面向对象方法设计的软件系统中,它的底层实现都是由N个对象组成的,所有的对象通过彼此的合作,最终实现系统的业务逻辑. 图1:软件系统中耦合的对象 如果我们打开机械 ...
- JFinal之ActiveRecord开发示例
JFinal独创Db + Record模式示例 JFinal配备的ActiveRecord插件,除了实现了类似Rails ActiveRecrod的功能之外,还实现了Db + Record模式,此模式 ...
- 找出1小时内占用cpu最多的10个进程的shell脚本
cpu时间是一项重要的资源,有时,我们需要跟踪某个时间内占用cpu周期最多的进程.在普通的桌面系统或膝上系统中,cpu处于高负荷状态也许不会引发什么问题.但对于需要处理大量请求的服务器来讲,cpu是极 ...
- 利用ajax,js以及正则表达式来验证表单递交
<!DOCTYPE html><html lang="en"> <head> <meta charset="utf-8" ...
- java列表组件鼠标双击事件的实现
Swing中提供两种列表组件,分别是列表框(JList)和组合框(JComboBox). 一.JList组件 构造方法: public JList():构造一个空的.具有只读模型的JList.publ ...
- java中关于string类和常量池的一点猜想
public class StringTest { /** * @param args */ public static void main(String[] args) { test1 ...
- shell脚本 screen管理
一.简介 源码地址 日期:2018/4/12 介绍:使用screen来启动程序,这个脚本可以管理screen 效果图: 二.使用 适用:centos6+ 语言:中文 注意:请先写一个脚本来启动java ...