题意:
解法:
最短路裸题.
唯一变化的点是边权是变化的,
对于x->(v,t,k),边权为(k-dist[x]%k)%k+t.
其他地方就是普通最短路,用dj算法求解即可.
code:
#include<bits/stdc++.h>
#define int long long
#define PI pair<int,int>
using namespace std;
const int maxm=2e6+5;
struct Node{
int v,t,k;
};
vector<Node>g[maxm];
int mark[maxm];
int dist[maxm];
int n,m,x,y;
void dj(int st){
for(int i=1;i<=n;i++)dist[i]=1e18;
dist[st]=0;
priority_queue<PI,vector<PI>,greater<PI> >q;
q.push({
dist[st],st});
while(q.size()){
int x=q.top().second;q.pop();
if(mark[x])continue;
mark[x]=1;
for(auto i:g[x]){
int v=i.v,t=i.t,k=i.k;
int time=(k-dist[x]%k)%k+t;
if(dist[v]>dist[x]+time){
dist[v]=dist[x]+time;
q.push({
dist[v],v});
}
}
}
}
void solve(){
cin>>n>>m>>x>>y;
for(int i=1;i<=m;i++){
int a,b,t,k;cin>>a>>b>>t>>k;
g[a].push_back({
b,t,k});
g[b].push_back({
a,t,k});
}
dj(x);
int ans=dist[y];
if(ans==1e18)ans=-1;
cout<<ans<<endl;
}
signed main(){
solve();
return 0;
}