E. Minimum spanning tree for each edge
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

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.

Input

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.

Output

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.

Sample test(s)
input
5 7
1 2 3
1 3 1
1 4 5
2 3 2
2 5 3
3 4 2
4 5 4
output
9
8
11
8
8
8
9

题意:给出一个图,问每一条边如果要在一个生成树当中,那这个生成树最小是多少。

分析:先找出一个最小生成树。

想像一下,加入一条边,会对这个生成树造成什么影响。

形成了一个环,然后最优情况,肯定要拿掉除他之外最大的一条边。

问题就变成了,在最小生成树上查询两点之间的边的最大值。

 /**
Create By yzx - stupidboy
*/
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <ctime>
#include <iomanip>
using namespace std;
typedef long long LL;
typedef double DB;
#define MIT (2147483647)
#define INF (1000000001)
#define MLL (1000000000000000001LL)
#define sz(x) ((int) (x).size())
#define clr(x, y) memset(x, y, sizeof(x))
#define puf push_front
#define pub push_back
#define pof pop_front
#define pob pop_back
#define mk make_pair inline int Getint()
{
int Ret = ;
char Ch = ' ';
bool Flag = ;
while(!(Ch >= '' && Ch <= ''))
{
if(Ch == '-') Flag ^= ;
Ch = getchar();
}
while(Ch >= '' && Ch <= '')
{
Ret = Ret * + Ch - '';
Ch = getchar();
}
return Flag ? -Ret : Ret;
} const int N = , M = ;
int n, m;
struct EdgeType
{
int u, v, value, index;
LL ans;
inline bool operator <(const EdgeType &t) const
{
return value < t.value;
} inline void Read()
{
u = Getint();
v = Getint();
value = Getint();
}
} edge[N];
int fa[N], favalue[N];
int first[N], to[N * ], value[N * ], next[N * ], tot;
int up[N][M], depth[N], maxcnt[N][M];
LL ans; inline void Input()
{
n = Getint();
m = Getint();
for(int i = ; i <= m; i++)
{
edge[i].Read();
edge[i].index = i;
}
} inline int Find(int x)
{
static int path[N], len;
for(len = ; x != fa[x]; x = fa[x])
path[++len] = x;
for(int i = ; i <= len; i++) fa[path[i]] = x;
return x;
} inline void Insert(int u, int v, int val)
{
tot++;
to[tot] = v, value[tot] = val, next[tot] = first[u];
first[u] = tot;
} inline void Bfs()
{
static int que[N], head, tail;
for(int i = ; i <= n; i++) fa[i] = -;
que[] = , head = tail = , fa[] = , depth[] = ;
while(head <= tail)
{
int u = que[head++];
for(int tab = first[u], v; tab; tab = next[tab])
if(fa[v = to[tab]] == -)
{
fa[v] = u, favalue[v] = value[tab], depth[v] = depth[u] + ;
que[++tail] = v;
}
}
} inline int GetMax(int u, int v)
{
int ret = , level = M;
while(depth[u] != depth[v])
{
if(depth[u] < depth[v]) swap(u, v);
while(level && ( << level) > depth[u] - depth[v]) level--;
ret = max(ret, maxcnt[u][level]);
u = up[u][level];
}
level = M;
while(level && u != v)
{
while(level && ( << level) > depth[u]) level--;
while(level && up[u][level] == up[v][level]) level--;
ret = max(ret, maxcnt[u][level]);
ret = max(ret, maxcnt[v][level]);
u = up[u][level], v = up[v][level];
}
while(u != v)
{
ret = max(ret, favalue[u]);
ret = max(ret, favalue[v]);
u = fa[u], v = fa[v];
}
return ret;
} inline bool CompareByIndex(const EdgeType &a, const EdgeType &b)
{
return a.index < b.index;
} inline void Solve()
{
sort(edge + , edge + + m);
for(int i = ; i <= n; i++) fa[i] = i;
for(int i = ; i <= m; i++)
{
int u = Find(edge[i].u), v = Find(edge[i].v);
if(u != v)
{
Insert(edge[i].u, edge[i].v, edge[i].value);
Insert(edge[i].v, edge[i].u, edge[i].value);
ans += edge[i].value;
fa[u] = v;
}
} Bfs(); for(int i = ; i < M; i++)
{
if(( << i) > n) break;
for(int j = ; j <= n; j++)
if(i == )
{
up[j][i] = fa[j];
maxcnt[j][i] = favalue[j];
}
else
{
up[j][i] = up[up[j][i - ]][i - ];
maxcnt[j][i] = max(maxcnt[j][i - ], maxcnt[up[j][i - ]][i - ]);
}
} for(int i = ; i <= m; i++)
{
int u = edge[i].u, v = edge[i].v;
int ret = GetMax(u, v);
edge[i].ans = ans - ret + edge[i].value;
} sort(edge + , edge + + m, CompareByIndex);
for(int i = ; i <= m; i++) printf("%I64d\n", edge[i].ans);
} int main()
{
freopen("a.in", "r", stdin);
Input();
Solve();
return ;
}

CF# Educational Codeforces Round 3 E. Minimum spanning tree for each edge的更多相关文章

  1. CF Educational Codeforces Round 3 E. Minimum spanning tree for each edge 最小生成树变种

    题目链接:http://codeforces.com/problemset/problem/609/E 大致就是有一棵树,对于每一条边,询问包含这条边,最小的一个生成树的权值. 做法就是先求一次最小生 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. Educational Codeforces Round 3 E. Minimum spanning tree for each edge (最小生成树+树链剖分)

    题目链接:http://codeforces.com/contest/609/problem/E 给你n个点,m条边. 问枚举每条边,问你加这条边的前提下组成生成树的权值最小的树的权值和是多少. 先求 ...

  7. [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 ...

  8. codeforces 609E 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 ...

  9. Codeforces Edu3 E. Minimum spanning tree for each edge

    time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standa ...

随机推荐

  1. C#索引器一

    索引器允许类或者结构的实例按照与数组相同的方式进行索引取值,索引器与属性类似,不同的是索引器的访问是带参的. 索引器和数组比较: (1)索引器的索引值(Index)类型不受限制 (2)索引器允许重载 ...

  2. NYOJ题目34韩信点兵

    aaarticlea/png;base64,iVBORw0KGgoAAAANSUhEUgAAAskAAAHiCAIAAACV1MbSAAAgAElEQVR4nO3dPXLjONeG4W8TyrUQx1 ...

  3. mysql入门语句10条

    1,连接数据库服务器 mysql  -h host   -u root   -p  xxx(密码) 2,查看所有库 show databases; 3,选库 use 库名 4,查看库下面的表 show ...

  4. 在ubuntu上搭建开发环境4---ubuntu简单的搭建LAMP环境和配置

    最近重新安装了Ubuntu,但是之前的LAMP环境自然也就没有了,实在是不想再去编译搭建LAMP环境(这种方法实在是太费时间,而且太容易遇到各种不知道为什么的错误),所以,就去查查有没有什么简单的搭建 ...

  5. Analysis Services OLAP 概述

    1. 什么是OLAP •定义1 :OLAP(联机分析处理)是针对特定问题的联机数据访问和分析.通过对信息(维数据)的多种可能的观察形式进行快速.稳定一致和交互性的存取,允许管理决策人员对数据进行深入观 ...

  6. 一致性hash算法简介与代码实现

    一.简介: 一致性hash算法提出了在动态变化的Cache环境中,判定哈希算法好坏的四个定义: 1.平衡性(Balance) 2.单调性(Monotonicity) 3.分散性(Spread) 4.负 ...

  7. oracle检查点队列(checkpoint queue)

    buffer cache CBC链 按地址链 LRU  干净buffer LRUW  脏buffer  按照冷热 checkpoint queue:链buffer,①链脏块②按buffer第一次脏的时 ...

  8. [读书笔记] Web 前端开发修炼之道

    原创地址:http://www.cnblogs.com/bnbqian/p/3735565.html 转载请注明出处 今天我们要读的书是Web 前端开发修炼之道 第1章 从网站重构说起 1.1 糟糕的 ...

  9. 让用VS2012/VS2013编写的程序在XP中顺利运行

    转自:http://blog.csdn.net/asanscape/article/details/38752655/ 微软为了推销自家平台,默认配置下VS2012和VS2013编写的应用程序只能在V ...

  10. fatal error C1083: 无法打开包括文件:“stddef.h”: No such file or directory

      在VS2012安装吗目录下的这个文件夹 D:\Program Files\Visual Studio 2012\VC\include 找到了stddef.h文件,将其加到项目的包含目录中,OK. ...