2017 Multi-University Training Contest - Team 4 phone call(树+lca+并查集)
题解:

(并查集处理往上跳的时候,一定要先让u,v往上跳到并查集的祖先,不然会wa掉)
代码如下:
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <cstring>
using namespace std;
const int maxn = 1e5 + ;
typedef long long LL;
int f[maxn], g[maxn], p[maxn], deep[maxn];
LL W[maxn];
int ffind(int x) { return f[x] == x ? f[x] : f[x] = ffind(f[x]); }
int gfind(int x) { return g[x] == x ? g[x] : g[x] = gfind(g[x]); }
struct Line{
int u1, v1, u2, v2;
int cost;
bool operator <(const Line& B) const{
return cost < B.cost;
}
};
vector<int> G[maxn];
vector<Line> V; void dfs(int x, int fa, int d){
deep[x] = d;
p[x] = fa;
for(int i = ; i < G[x].size(); i++){
int to = G[x][i];
if(to == fa) continue;
dfs(to, x, d+);
}
} void Merge(int u, int v, LL w){
u = ffind(u); v = ffind(v);
while(ffind(u) != ffind(v)){
if(deep[u] < deep[v]) swap(u, v);
int fa = ffind(u);
u = p[fa];
f[fa] = ffind(u);
u = ffind(u);
if(gfind(fa) != gfind(u)){
W[gfind(u)] += (W[gfind(fa)] + w);
g[gfind(fa)] = gfind(u);
}
}
} int main()
{
int T, n, m, x, y;
cin>>T;
while(T--){
cin>>n>>m;
memset(W, , sizeof(W));
for(int i = ; i <= n; i++) g[i] = f[i] = i;
for(int i = ; i <= n; i++) G[i].clear();
V.clear();
V.resize(m);
for(int i = ; i < n; i++){
scanf("%d %d", &x, &y);
G[x].push_back(y);
G[y].push_back(x);
}
dfs(, , );
for(int i = ; i < m; i++){
scanf("%d %d %d %d %d", &V[i].u1, &V[i].v1, &V[i].u2, &V[i].v2, &V[i].cost);
}
sort(V.begin(), V.end());
for(int i = ; i < V.size(); i++){
Line line = V[i];
int u = line.u1, v = line.v1, lca1, lca2;
Merge(u, v, line.cost);
lca1 = ffind(u);
u = line.u2, v = line.v2;
Merge(u, v, line.cost);
lca2 = ffind(u);
if(gfind(lca1) != gfind(lca2)) {
W[gfind(lca2)] += (W[gfind(lca1)] + line.cost);
g[gfind(lca1)] = gfind(lca2);
}
}
int num = ;
for(int i = ; i <= n; i++) if(gfind(i) == gfind()) num++;
cout<<num<<" "<<W[gfind()]<<endl;
}
return ;
}
2017 Multi-University Training Contest - Team 4 phone call(树+lca+并查集)的更多相关文章
- 2017 Multi-University Training Contest - Team 9 1005&&HDU 6165 FFF at Valentine【强联通缩点+拓扑排序】
FFF at Valentine Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) ...
- 2017 Multi-University Training Contest - Team 9 1004&&HDU 6164 Dying Light【数学+模拟】
Dying Light Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Tot ...
- 2017 Multi-University Training Contest - Team 9 1003&&HDU 6163 CSGO【计算几何】
CSGO Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Subm ...
- 2017 Multi-University Training Contest - Team 9 1002&&HDU 6162 Ch’s gift【树链部分+线段树】
Ch’s gift Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total S ...
- 2017 Multi-University Training Contest - Team 9 1001&&HDU 6161 Big binary tree【树形dp+hash】
Big binary tree Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)T ...
- 2017 Multi-University Training Contest - Team 1 1003&&HDU 6035 Colorful Tree【树形dp】
Colorful Tree Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)T ...
- 2017 Multi-University Training Contest - Team 1 1006&&HDU 6038 Function【DFS+数论】
Function Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total ...
- 2017 Multi-University Training Contest - Team 1 1002&&HDU 6034 Balala Power!【字符串,贪心+排序】
Balala Power! Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)T ...
- 2017 Multi-University Training Contest - Team 1 1011&&HDU 6043 KazaQ's Socks【规律题,数学,水】
KazaQ's Socks Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)T ...
随机推荐
- web前端逻辑计算,血的教训
在web前端进行页面开发的过程中,难免的遇到逻辑问题,这不是什么大问题,既然走上IT条黑道,那小伙伴们的逻辑推理能力及逻辑计算能力是不会有太大问题的. 然而,有的逻辑计算,就算你逻辑计算能力超强,也不 ...
- node.js中http通讯模块
创建一个服务器 首先建立一个js文件,命名为app.js写入内容: const http=require('http'); http.createServer((request,response)=& ...
- pyqt5--学习资料
http://zetcode.com/gui/pyqt5/ http://www.thehackeruniversity.com/2014/01/23/pyqt5-beginner-tutorial/ ...
- HTTP-Http状态码301和302的区别
官方解释: 301 redirect: 301 代表永久性转移(Permanently Moved) 302 redirect: 302 代表暂时性转移(Temporarily Moved ) 共同点 ...
- ReentrantLock详解
ReentrantLock概述 ReentrantLock是Lock接口的实现类,可以手动的对某一段进行加锁.ReentrantLock可重入锁,具有可重入性,并且支持可中断锁.其内部对锁的控制有两种 ...
- JavaScript实现判断图片是否加载完成的3种方法整理
JavaScript实现判断图片是否加载完成的3种方法整理 有时候我们在前端开发工作中为了获取图片的信息,需要在图片加载完成后才可以正确的获取到图片的大小尺寸,并且执行相应的回调函数使图片产生某种显示 ...
- atoi 和 atof (把数字字符串转化为数字储存)
int atoi(char *s) 如果字符串内容是整数就返回该整数,否则返回0 double atof(char *s) 同上,不过返回浮点型 #include<iostream> #i ...
- 程序员编程利器:20款最好的免费的IDEs和编辑器
程序员编程利器:20款最好的免费的IDEs和编辑器 还没转眼明年可就大年三十了,忙的可真是晕头转了个向,看着亲朋好友们那让人欣羡的小肚腩,不禁感慨,岁月是一把猪饲料,绿了芭蕉,肥了那杨柳小蛮腰,可怜我 ...
- 每天一个Linux命令(13):apt命令
apt-get和apt-cache命令是Debian Linux发行版中的APT软件包管理工具.所有基于Debian的发行都使用这个包管理系统.deb包可以把一个应用的文件包在一起,大体就如同Wind ...
- 基于Python的接口自动化-01
为什么要做接口测试 当前互联网产品迭代速度越来越快,由之前的2-3个月到个把月,再到班车制,甚至更短,每次发版之前都需要对所有功能进行回归测试,在人力资源有限的情况下,做自动化测试很有必要.由于UI更 ...