Codeforces 609E Minimum spanning tree for each edge

最新推荐文章于 2022-07-19 12:17:47 发布

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最新推荐文章于 2022-07-19 12:17:47 发布

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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.

题目大意：给你一个图，对于每条边，求出包含这条边的最小生成树的权值和。

思路：我们可以看出，对于每个最小生成树来说，它的主体其实还是整个图的最小生成树。对于每条边(u,v)，我们分情况讨论：若u,v直接连通，证明u->v这条边已经在最小生成树中了，直接输出答案；若u,v不直接连通，我们就选取u->v的路径上的最大边，去掉，再把u,v连通，输出答案。不难证明，去掉u->v路径上的一条边，再连通u,v后，图仍然是一棵树。故我们需要求两点之间边权的最大值，采用倍增LCA维护。

#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<cstdlib>
#include<ctime>

using namespace std;
const int MAXN=200010;
const int MAXLOG=20;
int n,m;
struct Edge
{
	int u,v,w,id;
};
Edge map[MAXN];
int cmp(const Edge a,const Edge b)
{
	return a.w<b.w;
}
int cmp1(const Edge a,const Edge b)
{
	return a.id<b.id;
}
struct edge
{
	int next,to,val;
};
edge e[MAXN*2];
int head[MAXN],cnt;
void addedge(int u,int v,int w)
{
	e[++cnt].next=head[u];
	e[cnt].to=v;
	e[cnt].val=w;
	head[u]=cnt;
}
int fa[MAXN];
int find(int x)
{
	if (fa[x]!=x)
	{
		fa[x]=find(fa[x]);
	}
	return fa[x];
}
long long ans;
bool use[MAXN];
void Kruskal()
{
	sort(map+1,map+1+m,cmp);
	for (int i=1;i<=n;i++)
	{
		fa[i]=i;
	}
	for (int i=1;i<=m;i++)
	{
		int f1=find(map[i].u),f2=find(map[i].v);
		if (f1==f2)
		{
			continue;
		}
		fa[f1]=f2;
		use[map[i].id]=true;
		ans+=map[i].w;
		addedge(map[i].u,map[i].v,map[i].w);
		addedge(map[i].v,map[i].u,map[i].w);
	}
}
int f[MAXN][MAXLOG],g[MAXN][MAXLOG];
int deep[MAXN]; 
bool visit[MAXN];
void dfs(int x)
{
	visit[x]=1;
	for (int i=head[x];i;i=e[i].next)
	{
		int v=e[i].to;
		if (visit[v])
		{
			continue;
		}
		f[v][0]=x;
		deep[v]=deep[x]+1;
		g[v][0]=e[i].val;
		dfs(v);
	}
}
void build()
{
	for (int j=1;j<MAXLOG;j++)
	{
		for (int i=1;i<=n;i++)
		{
			f[i][j]=f[f[i][j-1]][j-1];
			g[i][j]=max(g[i][j-1],g[f[i][j-1]][j-1]);
		}
	}
}
int maxx;
void go_up(int &x,int d)
{
	for (int i=0;i<MAXLOG;i++)
	{
		if (d&(1<<i))
		{
			maxx=max(g[x][i],maxx);
			x=f[x][i];
		}
	}
}
void LCA(int x,int y)
{
	if (deep[x]>deep[y])
	{
		go_up(x,deep[x]-deep[y]);
	}
	else
	{
		go_up(y,deep[y]-deep[x]);
	}
	if (x==y)
	{
		return ;
	}
	for (int i=MAXLOG-1;i>=0;i--)
	{
		if (f[x][i]!=f[y][i])
		{
			maxx=max(maxx,max(g[x][i],g[y][i]));
			x=f[x][i];
			y=f[y][i];	
		}	
	} 
	maxx=max(maxx,max(g[x][0],g[y][0]));
	return ;
}
int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++)
	{
		scanf("%d%d%d",&map[i].u,&map[i].v,&map[i].w);
		map[i].id=i;
	}
	Kruskal();
	deep[1]=1;
	dfs(1);
	build();
	sort(map+1,map+1+m,cmp1);
	for (int i=1;i<=m;i++)
	{
		if (use[i])
		{
			printf("%I64d\n",ans);
		}
		else
		{
			maxx=0;
			LCA(map[i].u,map[i].v);
			printf("%I64d\n",(long long)ans-maxx+map[i].w);
		}
	}
    return 0;
}