- #include
- using namespace std;
- using VI = vector<int>;
- using ll = long long;
- using PII = pair <int , int>;
- const int mod = 998244353;
- int n,m,d;
- int idx = 0 ;
- struct edge{
- int to,from;
- int next;
- int v,l;
- }g[200010];
- int head[200010];
- void add(int a , int b , int v , int l){
- idx++;
- g[idx].from = a;
- g[idx].to = b;
- g[idx].v = v;
- g[idx].l = l;
- g[idx].next = head[a];
- head[a] = idx;
- }
- double dis[200][510];//达到某个点的时间 , 速度为j
- int vis[200][510];
- PII fa[200][510];
- double calc(int x, int y){
- return x * 1.0 / (y * 1.0);
- }
- int finalV;
-
-
- void dij(int s){
- //时间,速度,点
- priority_queue
double,PII>> pq; - memset(dis , 126 ,sizeof dis);
- //printf("%lf\n" , dis[2][25]);
- dis[1][70] = 0;
- fa[1][70] = {0,0};
- pq.push({0,{70,1}});
- while(pq.size()){
- auto x = pq.top().second;
- pq.pop();
- int v = x.first;
- int now = x.second;
- if(vis[now][v]) continue;
- vis[now][v] = 1;
- for(int i = head[now] ; i ; i = g[i].next){
- int to = g[i].to;
- int l = g[i].l;
- if(g[i].v == 0){
- if(dis[to][v] > dis[now][v] + calc(l,v)){
- dis[to][v] = dis[now][v] + calc(l,v);
- fa[to][v] = {now , v};
- pq.push({-dis[to][v] ,{v,to}});
- }
- }else{
- if(dis[to][g[i].v] > dis[now][v] + calc(l ,g[i].v)){
- dis[to][g[i].v] = dis[now][v] + calc(l ,g[i].v);
- fa[to][g[i].v] = {now , v};
- pq.push({-dis[to][g[i].v],{g[i].v , to}});
- }
- }
-
- }
- }
- }
-
- int output(int p , int v){
- if(p == 0) return 0;
- output(fa[p][v].first , fa[p][v].second);
- printf("%d " , p - 1);
- return 0;
- }
-
-
-
-
- int main(){
- cin>>n>>m>>d;
- d++;
- for(int i = 1 ; i <= m ; i++){
- int a,b,v,l;
- cin>>a>>b>>v>>l;
- a++,b++;
- add(a,b,v,l);
- }
-
- dij(1);
- int t = -1;
- for(int i = 0 ; i <= 510 ; i++){
- if(t == -1 || dis[d][t] > dis[d][i]){
- t = i;
- }
- }
- //cout<
- //printf("%lf\n", dis[d][t]);
- //printf("%d\n", fa[d][t].first);
- //cout<
- output(d , t);
-
-
-
- }
其实并不是真的建立分层图,只是有分层图的思想,或者是dp思想在里面
dp[i][j] // 到达 i 点速度是 j 的花费
v == 0 dp[i][x] = min( dp[u][x] + calc(len[u][i] / x) )
v != 0 dp[i][v] = min( dp[u][x] + calc(len[u][i] / v) )