codeforces 609E Minimum spanning tree for each edge
2 seconds
256 megabytes
standard input
standard output
Connected undirected weighted graph without self-loops and multiple edges is given. Graph contains n vertices and m edges.
For each edge (u, v) find the minimal possible weight of the spanning tree that contains the edge (u, v).
The weight of the spanning tree is the sum of weights of all edges included in spanning tree.
First line contains two integers n and m (1 ≤ n ≤ 2·105, n - 1 ≤ m ≤ 2·105) — the number of vertices and edges in graph.
Each of the next m lines contains three integers ui, vi, wi (1 ≤ ui, vi ≤ n, ui ≠ vi, 1 ≤ wi ≤ 109) — the endpoints of the i-th edge and its weight.
Print m lines. i-th line should contain the minimal possible weight of the spanning tree that contains i-th edge.
The edges are numbered from 1 to m in order of their appearing in input.
5 7
1 2 3
1 3 1
1 4 5
2 3 2
2 5 3
3 4 2
4 5 4
9
8
11
8
8
8
9
保证某条边e存在的MST就是普通Kruskal把e优先到了最前面。
先求一遍MST,如果e不再MST上,是因为形成了环,把环上除了e的最大权边去掉就好了。
(以前的LCA:用ST来RMQ,查询O(1)
(向祖先结点倍增其实和ST差不多,查询O(logn),维护信息灵活
(一开始想的是树剖,复杂度稍高
#include<bits/stdc++.h>
using namespace std; typedef long long ll; const int N = 2e5+, M = N*; int pa[N], rak[N];
int fd(int x){ return pa[x] ? pa[x] = fd(pa[x]) : x; }
bool unite(int x,int y)
{
int a = fd(x), b = fd(y);
if(a == b) return false;
if(rak[a] < rak[b]){
pa[a] = b;
}
else {
pa[b] = a;
if(rak[a] == rak[b]) rak[a]++;
}
return true;
} int fro[N], to[N], we[N]; int hd[N];
int nx[M], ver[M], wei[M];
int ec; void add_e(int u,int v,int w)
{
ver[++ec] = v;
wei[ec] = w;
nx[ec] = hd[u];
hd[u] = ec;
} int n, m;
int *cmp_c;
bool cmp_id(int i,int j){ return cmp_c[i] < cmp_c[j]; } int r[N];
ll kruskal()
{
ll re = ;
int i,j;
for(i = ; i <= m; i++) r[i] = i;
cmp_c = we;
sort(r+, r + + m, cmp_id);
//ec = 0;
for(i = ; i <= m; i++){
j = r[i];
if(unite(fro[j],to[j])){
add_e(fro[j],to[j],we[j]);
add_e(to[j],fro[j],we[j]);
re += we[j];
we[j] = ;
}
}
return re;
} const int LOG = ; int fa[N][LOG], mx[N][LOG];
int dep[N]; void dfs(int u,int f = ,int fw = ,int d = )
{
fa[u][] = f;
mx[u][] = fw;
dep[u] = d;
for(int i = hd[u]; i; i = nx[i]) {
int v = ver[i];
if(v == f) continue;
dfs(v,u,wei[i],d+);
}
} int lg; int queryMx(int u,int v)
{
int re = , i;
if(dep[u] < dep[v]) swap(u,v);
for(i = lg; i >= ; i--) if(dep[u] - (<<i) >= dep[v]){
re = max(re,mx[u][i]);
u = fa[u][i];
}
if(u == v) return re;
for(i = lg; i >= ; i--) if(fa[u][i] != fa[v][i]){
re = max(re,max(mx[u][i],mx[v][i]));
u = fa[u][i];
v = fa[v][i];
}
return max(re,max(mx[u][],mx[v][]));
} //#define LOCAL
int main()
{
#ifdef LOCAL
freopen("in.txt","r",stdin);
#endif
//cout<<log2(N);
scanf("%d%d",&n,&m);
int i,j;
for(i = ; i <= m; i++){
scanf("%d%d%d",fro+i,to+i,we+i);
}
ll mst = kruskal(); dfs();
lg = ceil(log2(n));
for(j = ; j <= lg; j++){
for(i = ; i <= n; i++) if(fa[i][j-]){
fa[i][j] = fa[fa[i][j-]][j-];
mx[i][j] = max(mx[i][j-],mx[fa[i][j-]][j-]);
}
}
for(i = ; i <= m; i++) {
printf("%I64d\n",we[i]?mst + we[i] - queryMx(fro[i],to[i]):mst);
}
return ;
}
codeforces 609E Minimum spanning tree for each edge的更多相关文章
- [Educational Round 3][Codeforces 609E. Minimum spanning tree for each edge]
这题本来是想放在educational round 3的题解里的,但觉得很有意思就单独拿出来写了 题目链接:609E - Minimum spanning tree for each edge 题目大 ...
- codeforces 609E. Minimum spanning tree for each edge 树链剖分
题目链接 给一个n个节点m条边的树, 每条边有权值, 输出m个数, 每个数代表包含这条边的最小生成树的值. 先将最小生成树求出来, 把树边都标记. 然后对标记的边的两个端点, 我们add(u, v), ...
- Educational Codeforces Round 3 E (609E) Minimum spanning tree for each edge
题意:一个无向图联通中,求包含每条边的最小生成树的值(无自环,无重边) 分析:求出这个图的最小生成树,用最小生成树上的边建图 对于每条边,不外乎两种情况 1:该边就是最小生成树上的边,那么答案显然 2 ...
- cf 609E.Minimum spanning tree for each edge
最小生成树,lca(树链剖分(太难搞,不会写)) 问存在这条边的最小生成树,2种情况.1.这条边在原始最小生成树上.2.加上这条半形成一个环(加上),那么就找原来这条边2端点间的最大边就好(减去).( ...
- Codeforces Educational Codeforces Round 3 E. Minimum spanning tree for each edge LCA链上最大值
E. Minimum spanning tree for each edge 题目连接: http://www.codeforces.com/contest/609/problem/E Descrip ...
- Codeforces Educational Codeforces Round 3 E. Minimum spanning tree for each edge 树上倍增
E. Minimum spanning tree for each edge 题目连接: http://www.codeforces.com/contest/609/problem/E Descrip ...
- Educational Codeforces Round 3 E. Minimum spanning tree for each edge LCA/(树链剖分+数据结构) + MST
E. Minimum spanning tree for each edge Connected undirected weighted graph without self-loops and ...
- CF# Educational Codeforces Round 3 E. Minimum spanning tree for each edge
E. Minimum spanning tree for each edge time limit per test 2 seconds memory limit per test 256 megab ...
- Educational Codeforces Round 3 E. Minimum spanning tree for each edge 最小生成树+树链剖分+线段树
E. Minimum spanning tree for each edge time limit per test 2 seconds memory limit per test 256 megab ...
随机推荐
- nginx 地址重写
例如, www.baidu.com 跳到 www.baidu.com/index.html #if ( $http_host ~* "^(.*)\.baidu\.com$" ...
- mysql 问题总结[转]
一.Can't connect to MySQL server on 'localhost' (10061) 不能连接到 localhost 上的mysql分析:这说明“localhost”计算机 ...
- (转)图解SSH原理
图解SSH原理 原文:https://www.jianshu.com/p/33461b619d53 http://blog.51cto.com/forlinux/1352900---------SSH ...
- 移动端本地 H5 秒开方案探索与实现
欢迎大家前往腾讯云+社区,获取更多腾讯海量技术实践干货哦~ 企业微信移动端项目中有需求要展示数据趋势的可视化图表,经过调研,最终决定以单页面 H5 来完成,对 APP 里的一些使用 H5 实现的功能模 ...
- mc:Ignorable="d"什么意思?
有两个命名空间我们要注意一下的:xmlns:d="http://schemas.microsoft.com/expression/blend/2008"xmlns:mc=" ...
- 初识Socket通信:基于TCP和UDP协议学习网络编程
学习笔记: 1.基于TCP协议的Socket网络编程: (1)Socket类构造方法:在客户端和服务器端建立连接 Socket s = new Socket(hostName,port);以主机名和端 ...
- Linux 连接 Xshell 及网络配置
一.准备工具 在WMware上已经装有Linux系统:WMware安装CentOS7文章. xshell连接工具: 二.修改相关配置 切换到root用户下: 配置主机名(可选): #方法一:替换原主机 ...
- [SQL SERVER系列]工作经常使用的SQL整理,实战篇(一)[原创]
工作经常使用的SQL整理,实战篇,地址一览: 工作经常使用的SQL整理,实战篇(一) 工作经常使用的SQL整理,实战篇(二) 工作经常使用的SQL整理,实战篇(三) 目录概览: 1.数据库 2.表 3 ...
- 服务器LIUNX之如何解决矿机问题
点进来的基本都是遇到liunx变矿机的小伙伴吧(cpu运载300%) 卡的连终端都很难打开 开下来之后提示 大意是, 到xxx网站给钱了事, 不过基本这个网站基本也上不去, 要么是暴力破解, 要么是通 ...
- VMware 安装提示缺少MicrosoftRuntime DLL 问题解决办法
VMware 安装提示缺少MicrosoftRuntime DLL 问题解决办法 刚刚安装VMware失败了试了好多办法,在这总结一下. 下面是程序的截图 这是报错信息 网上的解决方法: 当出现安装失 ...