CF609E. Minimum spanning tree for each edge

题解：随便构造一颗最小生成树然后对于其他不在树上的边考虑到删除这条链上的最大值在把这条边加上去能得到这条边所在的最小生成树可以LCT维护但是明显这个题是静态的树就没必要LCT 当然我觉得最优的是树剖以后ST nlogn的的复杂度也可以树剖+线段树nlog^2的复杂度

#include <bits/stdc++.h>
const int MAXN=2e5+10;
#define ll long long
using namespace std;
ll read(){  
    ll x=0,f=1;char ch=getchar();  
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}  
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();  
    return f*x;  
}
typedef struct Edge{
	int u,v,vul,id;
	friend bool operator<(Edge aa,Edge bb){
		return aa.vul<bb.vul;
	}
}Edge;
Edge ed[MAXN];int mu[MAXN],ma[21];
int f[MAXN];ll ans[MAXN];
int n,m;
int find1(int x){
	if(f[x]==x)return x;
	return f[x]=find1(f[x]);
}
vector<Edge>v1;
vector<Edge>v2;
vector<pair<int,int> >vec[MAXN];
int fa[MAXN],num[MAXN],dep[MAXN],son[MAXN];
int dp[MAXN][21],vul[MAXN];
void dfs(int v,int pre,int deep){
	fa[v]=pre;num[v]=1;dep[v]=deep+1;
	for(int i=0;i<vec[v].size();i++){
		if(vec[v][i].first!=pre){
			//F[vec[v][i].first][0]=pre;
			vul[vec[v][i].first]=vec[v][i].second;
			dfs(vec[v][i].first,v,deep+1);
			num[v]+=num[vec[v][i].first];
			if(son[v]==-1||num[vec[v][i].first]>num[son[v]])son[v]=vec[v][i].first;
		}
	}
}
int tp[MAXN],p[MAXN],fp[MAXN],cnt;
void dfs1(int v,int td){
	tp[v]=td;p[v]=++cnt;fp[p[v]]=v;
	if(son[v]!=-1)dfs1(son[v],td);
	for(int i=0;i<vec[v].size();i++){
		if(vec[v][i].first!=fa[v]&&vec[v][i].first!=son[v])dfs1(vec[v][i].first,vec[v][i].first);
	}
}
void st(){
	for(int i=1;i<=n;i++)dp[i][0]=vul[fp[i]];
	for(int j=1;ma[j]<=n;j++){
		for(int i=1;i<=n;i++){
			dp[i][j]=max(dp[i][j-1],dp[i+ma[j-1]][j-1]);
		}
	}
}
int rmq(int l,int r){
	int k=mu[r-l+1];
	return max(dp[l][k],dp[r-ma[k]+1][k]);
}
int Lca(int u,int v){
	int maxx=-1;
	int uu=tp[u];int vv=tp[v];
	while(uu!=vv){
		if(dep[uu]<dep[vv])swap(uu,vv),swap(u,v);
		maxx=max(maxx,rmq(p[uu],p[u]));
		u=fa[uu];uu=tp[u];
	}
	if(dep[u]>dep[v])swap(u,v);
	if(u!=v)maxx=max(maxx,rmq(p[son[u]],p[v]));
	return maxx;
}
int main(){
	ma[0]=1;
	for(int i=1;i<21;i++)ma[i]=ma[i-1]<<1;
	mu[0]=-1;
	for(int i=1;i<MAXN;i++){
    if((i&(i-1))==0)mu[i]=mu[i-1]+1;
    else mu[i]=mu[i-1];
    }
	n=read();m=read();
	for(int i=1;i<=n;i++)f[i]=i,son[i]=-1;
	for(int i=1;i<=m;i++)ed[i].id=i,ed[i].u=read(),ed[i].v=read(),ed[i].vul=read();
	sort(ed+1,ed+m+1);ll sum=0;
	for(int i=1;i<=m;i++){
		int t1=find1(ed[i].u);int t2=find1(ed[i].v);
		if(t1!=t2){
			f[t1]=t2;sum+=ed[i].vul;
			v2.push_back(ed[i]);
			vec[ed[i].u].push_back(make_pair(ed[i].v,ed[i].vul));
			vec[ed[i].v].push_back(make_pair(ed[i].u,ed[i].vul));
		}
		else v1.push_back(ed[i]);
	}
	dfs(1,0,0);dfs1(1,1);
	st();
	for(int i=0;i<v2.size();i++)ans[v2[i].id]=sum;
	for(int i=0;i<v1.size();i++){
		int key=Lca(v1[i].u,v1[i].v);
	//	cout<<key<<endl;
		ans[v1[i].id]=sum-key+v1[i].vul;
	}
	for(int i=1;i<=m;i++)printf("%lld\n",ans[i]);
	return 0;
}

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.

Examples

input

Copy

output

Copy

CF609E. Minimum spanning tree for each edge

猜你喜欢