AtCoder Beginner Contest 204

E - Rush Hour 2
設函式f(t)=t+ci+di/(t+1);
易得當t=sqrt(di)-1時取最小
所以根據時間來做dij
當時間大於sqrt(di)-1時直接加入即可
同時用並查集來檢視1和n是否聯通即可
ac code：

點選檢視程式碼

#include<bits/stdc++.h>
#define int long long
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define pb push_back
#define pii pair<int,int>
using namespace std;
const int mod=998244353;
const int maxn=500050;
int n,m,s,cnt;
const int inf=1e18;
int dis[maxn],h[maxn],to[maxn],c[maxn],nxt[maxn],vis[maxn],d[maxn];
struct node{
    int v,w;
    friend bool operator <(node a,node b){
        return a.w>b.w;
    }
};
priority_queue<node>q;
int f[maxn];
int find(int x){
    return f[x]==x?x:f[x]=find(f[x]);
}
void add(int a,int b,int c1,int d1){
    to[++cnt]=b;
    c[cnt]=c1;
    d[cnt]=d1;
    nxt[cnt]=h[a];
    h[a]=cnt;
}
void dijkstra(){
    for(int i=0;i<=n;i++)dis[i]=inf;
    dis[s]=0;
    node tmp;
    tmp.v=s;tmp.w=0;q.push(tmp);
    while(!q.empty()){
        int u=q.top().v,nowt=q.top().w;q.pop();
        if(vis[u])continue;vis[u]=1;
        for(int i=h[u];i;i=nxt[i]){
            int t=nowt;
            if(t<=round(sqrt(d[i]))-1)t=round(sqrt(d[i]))-1;
            if(dis[to[i]]>t+c[i]+(d[i]/(t+1))){
                dis[to[i]]=t+c[i]+(d[i]/(t+1));
                tmp.w=dis[to[i]];
                tmp.v=to[i];
                q.push(tmp);
            }
        }
    }
}
void solve(){
    cin>>n>>m;
    s=1;
    for(int i=1;i<=n;i++)f[i]=i;
    for(int i=1,u,v,w,k;i<=m;i++){
        cin>>u>>v>>w>>k;
        add(u,v,w,k);add(v,u,w,k);
        f[find(u)]=find(v);
    }
    if(find(1)!=find(n)){cout<<-1;return;}
    dijkstra();
    cout<<dis[n];
}
signed main(){
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    int _=1;
//    cin>>_;
    while(_--)solve();
}