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
- DevOps - 配置管理 - Ansible
http://www.zsythink.net/archives/category/运维相关/ansible/
- 转:Java子线程中的异常处理(通用)
引自:https://www.cnblogs.com/yangfanexp/p/7594557.html 在普通的单线程程序中,捕获异常只需要通过try ... catch ... finally . ...
- 6 大主流 Web 框架优缺点对比(转)
英文: Kit Kelly 译文:oschina https://www.oschina.net/translate/web-frameworks-conclusions 是该读些评论和做一些总结 ...
- 五、RegExp(正则表达式)篇
正则表达式,只用记住: 0./pattern/igm i--不区分大小写 g--找到所有相匹配的 m--多行匹配 可以只写其中一个 ps:/pattern/i (无视大小写) 1." ...
- XPath Helper的安装使用
XPath Helper的安装使用 xpath helper 是一款chrome浏览器插件,主要用来分析当前网页信息的xpath,在抓取数据时一般会使用到xpath. 安装 下载地址:http://c ...
- Python学习第二弹
昨天补充: 编码: Unicode ; utf-8 ; GBK 关系: 关键字:1. continue 终止当前循环,进行下一次循环 2. break 终止循环 题6解法2: ...
- 怎么用Python Flask模板jinja2在网页上打印显示16进制数?
问题:Python列表(或者字典等)数据本身是10进制,现在需要以16进制输出显示在网页上 解决: Python Flask框架中 模板jinja2的If 表达式和过滤器 假设我有一个字典index, ...
- EXKMP学习笔记QAQ
因为一本通少了一些算法,所以我就自行补充了一些东西上去. EXKMP也就是扩展KMP,是一种特别毒瘤的东西 EXKMP确实很难,我理解他的时间与AC机的时间差不多,而且还很难记,因此一学会就马上写博客 ...
- DHT11资料
产品名:温湿度传感器 型号:DHT11 厂商:奥松电子 参数: 相对湿度: 分辨率:0.1%RH 16Bit 精度:25℃ 正负 %2 温度: 分辨率:0.1%RH 16 ...
- linux文件操作篇 (四) 目录操作
#include <sys/stat.h>#include <unistd.h>#include <dirent.h> //创建文件夹 路径 掩码 int mkdi ...