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;
}
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.
Examples
input
Copy
5 71 2 31 3 11 4 52 3 22 5 33 4 24 5 4
output
Copy
98118889
原文地址：https://www.cnblogs.com/wang9897/p/9195843.html