uva 12655 Trucks [LCA](树链剖分+MST)
The Subtle Balloons Company (SBC) is the main balloon provider for programming contests; it has
huge factories and warehouses, as well as an extensive truck
eet to ensure the contestants' happiness.
There are lots of competition sites in Nlogonia, and all of them hired SBC for supplying balloons for
their contests. Nlogonia is an archipelago connected by several bridges. Every island of Nlogonia may
have several regional sites and may also house several SBC warehouses.
When planning the routes for balloon deliveries, SBC faced a problem: for safety issues, every
bridge in Nlogonia has some maximum weight limit for vehicles which cross it. And because of the
great net weight of the transported merchandise, SBC operations' chief asked you to write a program to
determine the maximum weight allowed to be transported between warehouses and competition sites.
Input
The input contains several test cases. The rst line of a test case contains three integers N, M and S
which indicate, respectively, the number of islands, the number of bridges that connect the islands and
the number of sites. The islands are numbered from 1 to N.
Each of the next M lines describes a bridge. The description of a bridge consists in a line with
three integers A, B and W, indicating respectively the two islands connected by the bridge and the
maximum weight allowed in that bridge, in tons.
All bridges are two-way roads; every pair of islands is connected by at most one bridge; and it is
possible to reach every other island in the archipelago using only bridges (naturally it may be needed
to pass through other islands to do so).
Each of the next S lines describe a competition site and contains two integers L and H indicat-
ing, respectively, the number of the island where this site is and the number of the island where the
wharehouse which will be used to deliver the balloons to the site is.
Output
For each site in a test case, in the order they were given, your program must produce a single line,
containing a single integer, the biggest weight which can be transported by truck from the warehouse
to the site.
Restrictions
• 2 ≤ N ≤ 2 × 104
• 1 ≤ M ≤ 105
• 1 ≤ S ≤ 5 × 104
• 1 ≤ A, B, L, H ≤ N, A ̸= B, L ̸= H
• 0 ≤ W ≤ 105
Sample Input
4 5 4
1 2 9
1 3 0
2 3 8
2 4 7
3 4 4
1 4
2 1
3 1
4 3
4 5 2
1 2 30
2 3 20
3 4 10
4 1 40
2 4 50
1 3
1 2
Sample Output
7
9
8
7
20
40
1: 如果(u,v)为轻边,则size(v)<=size(u)/2;
2: 从根到某一点的路径上轻边的个数不大于O(logN)
时间复杂度O(N*logN*logN)
树链剖分:


view code#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long ll;
const int INF = 1<<30;
const int M = 200010;
const int N = 20010;
int n, m, Q;
int pre[N], sz[N], fa[N], top[N], id[N], fid[N];
int dep[N], gid, son[N];
int Min[N<<2]; struct edge
{
int u, v, w, next;
bool operator < (const edge &o)const{
return w>o.w;
}
edge(int u, int v, int w):u(u),v(v),w(w) {}
edge() {}
edge(int u, int v, int w, int next):u(u),v(v),w(w),next(next) {}
}e[M], mst[M];
int mcnt; void addedge(int u, int v, int w)
{
mst[mcnt] = edge(u, v, w, pre[u]);
pre[u] = mcnt++;
mst[mcnt] = edge(v, u, w, pre[v]);
pre[v] = mcnt++;
} int find(int x)
{
return x==fa[x]?x:(fa[x]=find(fa[x]));
} void MST()
{
mcnt = 0;
for(int i=0; i<=n; i++) fa[i] = i;
memset(pre, -1, sizeof(pre));
sort(e, e+m);
for(int i=0; i<m; i++)
{
int u = find(e[i].u), v = find(e[i].v);
if(u==v) continue;
// printf("(%d, %d) ->%d\n", e[i].u, e[i].v, e[i].w);
fa[u] = v;
addedge(e[i].u, e[i].v, e[i].w);
}
} void dfs(int u, int f, int d)
{
// fa[u]表示u的父亲,dep[u]表示u的深度
//sz[u]表示u字节点的个数
//son[u]与u在同重链上的儿子节点
fa[u] = f; dep[u] = d; son[u] = 0; sz[u] = 1;
for(int i=pre[u]; ~i; i=mst[i].next)
{
int v = mst[i].v;
if(v==f) continue;
dfs(v, u, d+1);
sz[u] += sz[v];
if(sz[son[u]] < sz[v]) son[u] = v;
}
} void get_pos(int u, int f)
{
//id[u]表示u与其父亲节点的连边,在线段树中的位置
id[u] = ++gid;
fid[gid] = u;//fid与id数组相反,在这道题没什么用
top[u] = f;//top[u] 表示u所在的重链的顶端节点
if(son[u]!=0) get_pos(son[u], f);
for(int i=pre[u]; ~i; i=mst[i].next)
{
int v = mst[i].v;
if(v==fa[u] || v==son[u]) continue;
get_pos(v, v);
}
} void build(int l, int r, int rt)
{
if(l==r){
Min[rt] = INF;
return ;
}
int m = (l+r)>>1;
build(lson);
build(rson);
} void update(int p, int c, int l, int r, int rt)
{
if(l==r){
Min[rt] = c;
return ;
}
int m = (l+r)>>1;
if(p<=m) update(p, c, lson);
else update(p, c, rson);
Min[rt] = min(Min[rt<<1], Min[rt<<1|1]);
} int query(int L, int R, int l ,int r, int rt)
{
if(L<=l && R>=r) return Min[rt];
int m = (l+r)>>1, ans = INF;
if(L<=m) ans = min(ans, query(L,R,lson));
if(R>m) ans = min(ans, query(L,R,rson));
return ans;
} int lca(int u, int v)
{
int fv = top[v], fu = top[u];
int ans = INF;
while(fv!=fu)
{
if(dep[fv]>dep[fu])
{
swap(fv, fu); swap(u,v);
}
ans = min(ans, query(id[fu], id[u], 1, gid, 1));
u = fa[fu];
fu = top[u];
}
if(dep[u]<dep[v]) swap(u, v);
if(u!=v) ans = min(ans, query(id[v]+1, id[u], 1, gid, 1));//ti[v]指的是v与其父亲的边,所以+1
return ans;
} int main()
{
// freopen("in.txt", "r", stdin);
while(scanf("%d%d%d", &n, &m, &Q)>0)
{
for(int i=0; i<m; i++)
scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w);
MST();
sz[1] = 0, gid = 0;
dfs(1, 1, 1);
get_pos(1, 1);
build(1, gid, 1);
int u, v;
for(int i=0; i<mcnt; i++)
{
int u = mst[i].u, v = mst[i].v;
if(dep[u]<dep[v]) swap(u,v);
update(id[u], mst[i].w, 1, gid, 1);
}
while(Q--)
{
scanf("%d%d", &u, &v);
printf("%d\n", lca(u,v));
}
}
return 0;
}
uva 12655 Trucks [LCA](树链剖分+MST)的更多相关文章
- Count on a tree SPOJ 10628 主席树+LCA(树链剖分实现)(两种存图方式)
Count on a tree SPOJ 10628 主席树+LCA(树链剖分实现)(两种存图方式) 题外话,这是我第40篇随笔,纪念一下.<( ̄︶ ̄)↗[GO!] 题意 是说有棵树,每个节点上 ...
- [BZOJ3626] [LNOI2014]LCA(树链剖分)
[BZOJ3626] [LNOI2014]LCA(树链剖分) 题面 给出一棵N个点的树,要求支持Q次询问,每次询问一个点z与编号为区间[l,r]内的点分别求最近公共祖先得到的最近公共祖先深度和.N, ...
- BZOJ 3626: [LNOI2014]LCA [树链剖分 离线|主席树]
3626: [LNOI2014]LCA Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 2050 Solved: 817[Submit][Status ...
- Codeforces Round #329 (Div. 2) D. Happy Tree Party LCA/树链剖分
D. Happy Tree Party Bogdan has a birthday today and mom gave him a tree consisting of n vertecie ...
- BZOJ 3626: [LNOI2014]LCA( 树链剖分 + 离线 )
说多了都是泪啊...调了这么久.. 离线可以搞 , 树链剖分就OK了... -------------------------------------------------------------- ...
- [CodeVS2370] 小机房的树 (LCA, 树链剖分, LCT)
Description 小机房有棵焕狗种的树,树上有N个节点,节点标号为0到N-1,有两只虫子名叫飘狗和大吉狗,分居在两个不同的节点上.有一天,他们想爬到一个节点上去搞基,但是作为两只虫子,他们不想花 ...
- BZOJ3626[LNOI2014]LCA——树链剖分+线段树
题目描述 给出一个n个节点的有根树(编号为0到n-1,根节点为0).一个点的深度定义为这个节点到根的距离+1.设dep[i]表示点i的深度,LCA(i,j)表示i与j的最近公共祖先.有q次询问,每次询 ...
- bzoj 3626 : [LNOI2014]LCA (树链剖分+线段树)
Description 给出一个n个节点的有根树(编号为0到n-1,根节点为0).一个点的深度定义为这个节点到根的距离+1.设dep[i]表示点i的深度,LCA(i,j)表示i与j的最近公共祖先.有q ...
- LCA树链剖分
LCA(Lowest Common Ancestor 最近公共祖先)定义如下:在一棵树中两个节点的LCA为这两个节点所有的公共祖先中深度最大的节点. 比如这棵树 结点5和6的LCA是2,12和7的LC ...
随机推荐
- IOS 之 PJSIP 笔记(二) iPJSUA 的简单使用
上一篇在编译完之后,就很不负责的结束了,本篇就对 PJSIP 库中提供的一个示例 iPJSUA 的使用,做一个简单的介绍.也能解除很多人对官方文档的一个困扰,起码我是被困扰过了. 首先,要确保你的 P ...
- WPF ListView 选中问题
WPF ListView 选中问题 摘自:http://www.cnblogs.com/BBHor/archive/2013/04/28/VisualTreeHelper-PreviewMouseD ...
- 无废话WCF入门教程一[什么是WCF]
http://www.cnblogs.com/iamlilinfeng/archive/2012/09/25/2700049.html wcf技术交流,同学习共进步,欢迎加群: 群号:3981831 ...
- ASP.NET MVC4 传递Model到View
原文发表在:http://www.star110.com/Note/ReadArticle/60641215331146140043.html 开发环境:.NET MVC4 + EF6.0 模型: 1 ...
- ExecutorService常用方法和newFixedThreadPool创建固定大小的线程池
1.ExecutorService: 是一个接口,继承了Executor: public interface ExecutorService extends Executor { } 2.Execut ...
- Linux Shell系列教程之(十七) Shell文件包含
本文是Linux Shell系列教程的第(十七)篇,更多Linux Shell教程请看:Linux Shell系列教程 通过文件包含,可以引用其他文件的内容,也可以将复杂内容分开,使程序结构更加清晰. ...
- expect入门--自动化linux交互式命令
很多linux程序比如passwd,ftp,scp,ssh等自身并没有提供一种静默式的执行选项,而是依赖于运行时的终端输入来进行后一步的操作比如更改密码.文件上传.下载等.虽然有些编程语言如java嵌 ...
- [mysql] timestamp自动更新和初始化
1.概述 在我们设计表的时候,考虑将行数据的创建时间和最后更新时间记录下来是很好的实践.尤其是可能需要做数据同步或者对数据新鲜度有要求的表.举些应用场景,更新距上次更新超过2小时的行数据,或者是将一个 ...
- jQuery.Cookie.js用法
jQuery.Cookie.js:一个轻量级的cookie插件,可以读取.写入.删除cookie. 一.使用方法 引入jQuery与jQuery.Cookie.js插件 <script src= ...
- ABAP中使用浏览器打开网页
在SAP ABAP中可以在Screen中嵌入Html control打开网页,也可以通过调用本地的IE浏览器打开. 1.在Screen中嵌入Html control的例子,在系统中有,se38:SAP ...