# P2934 [usaco09jan] safe travel g (parallel search & shortest path tree)

2021-08-10 08:44:06

P2934 [USACO09JAN]Safe Travel G( Union checking set & Shortest path tree )

Ideas

The last updated edge of each node in the shortest path of a graph can form a shortest path spanning tree .

According to the subject conditions , We want to find the shortest path of each node without passing through the parent side .

Consider for the current node i i i First pass through a node of its subtree u u u, Then pass by a non tree u → v u\rightarrow v uv, And then from v → 1 v\rightarrow 1 v1.

The answer to this calculation is ： d u + d v + w u , v − d i d_u+d_v+w_{u,v}-d_i du+dv+wu,vdi, Note that the first three items are related to i i i irrelevant , So we can find all non tree edges first , Then sort , Then update the answer with these edges , Which nodes can be updated for each edge ？

Obviously, it can be updated u → l c a ( u , v ) u\rightarrow lca(u,v) ulca(u,v) and v → l c a ( u , v ) v\rightarrow lca(u,v) vlca(u,v) All nodes on these two paths , barring l c a lca lca In itself . In other words, the two endpoints of this edge are not in the same subtree .

We only need to update each node once , Because the front edge must be better than the back edge after sorting .

code

// Problem: P2934 [USACO09JAN]Safe Travel G
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P2934
// Memory Limit: 125 MB
// Time Limit: 1000 ms
// Date: 2021-04-01 11:47:06
// --------by Herio--------

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N=1e5+5,M=2e5+5,inf=0x3f3f3f3f,mod=1e9+7;
#define mst(a,b) memset(a,b,sizeof a)
#define PII pair<int,int>
#define fi first
#define se second
#define pb emplace_back
#define SZ(a) (int)a.size()
#define IOS ios::sync_with_stdio(false),cin.tie(0)
void Print(int *a,int n){
for(int i=1;i<n;i++)
printf("%d ",a[i]);
printf("%d\n",a[n]);
}
int n,m,h[N],cnt,d[N];
int s[N];
int f[N];
int ans[N];
int find(int x){return x==s[x]?x:s[x]=find(s[x]);}
void Init(int n){
for(int i=1;i<=n;i++) s[i]=i,ans[i]=-1;
}
struct edge{
int to,nt,w;
}e[M<<1];
e[++cnt]={v,h[u],w},h[u]=cnt;
e[++cnt]={u,h[v],w},h[v]=cnt;
}
struct node{
int u,v,w;
}a[M],b[M];
int tot;
void dij(){
priority_queue<PII>q;
mst(d,0x3f);d=0;
q.push({0,1});
while(!q.empty()){
int u=q.top().se,l=-q.top().fi;q.pop();
if(l>d[u]) continue;
for(int i=h[u];i;i=e[i].nt){
int v=e[i].to,w=e[i].w;
if(d[v]>d[u]+w) f[v]=u,d[v]=d[u]+w,q.push({-d[v],v});
}
}
}
bool cmp(node a,node b){
return a.w<b.w;
}
void fun(node &a){
int u=a.u,v=a.v,w=a.w;
while(find(u)!=find(v)){
if(d[find(u)]<d[find(v)]) swap(u,v);
ans[find(u)]=w-d[find(u)];
u=s[find(u)]=f[find(u)];
}
}
int main(){
scanf("%d%d",&n,&m);
for(int i=1,u,v,w;i<=m;i++){
scanf("%d%d%d",&u,&v,&w);
a[i]={u,v,w};
}
dij();
for(int i=1;i<=m;i++){
int u=a[i].u,v=a[i].v,w=d[u]+d[v]+a[i].w;
if(f[u]!=v&&f[v]!=u) b[++tot]={u,v,w};
}
sort(b+1,b+tot+1,cmp);
Init(n);
for(int i=1;i<=tot;i++)
fun(b[i]);
for(int i=2;i<=n;i++) printf("%d\n",ans[i]);
return 0;
}


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