传送门啦
看起来是一个最短路问题,但是引入了速度限制,就要写一下二维最短路了。
$ dis[i][j] $ :表示到i这个点,速度为j的最短时间。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn = 600;
const int maxm = 200050;
inline int read(){
char ch = getchar();
int f = 1 , x = 0;
while(ch > '9' || ch < '0'){if(ch == '-')f = -1;ch = getchar();}
while(ch >= '0' && ch <= '9'){x = (x << 1) + (x << 3) + ch - '0';ch = getchar();}
return x * f;
}
int n,m,d,a,b,v,l;
int head[maxn],tot;
double dis[maxn][maxn];
bool inq[maxn][maxn];
struct Edge {
int from,to,val,next,len;
}edge[maxm << 1];
inline void add(int u,int v,int sp,int l){
edge[++tot].from = u;
edge[tot].to = v;
edge[tot].val = sp;
edge[tot].len = l;
edge[tot].next = head[u];
head[u] = tot;
}
struct Node {int u,v;};
Node pre[maxn][maxn];
inline void spfa(){
queue<Node> q;
q.push((Node) {0 , 70});
for(int i=0;i<=n;i++)
for(int j=0;j<=500;j++){
dis[i][j] = 1e9;
pre[i][j].u = pre[i][j].v = -1;
}
dis[0][70] = 0 , inq[0][70] = 1;
while(!q.empty()){
Node cur = q.front();
q.pop();
int u = cur.u , v = cur.v;
inq[u][v] = 0;
for(int i=head[u];i;i=edge[i].next){
int to = edge[i].to;
int vv = edge[i].val ? edge[i].val : v;
if(dis[to][vv] > dis[u][v] + (double)edge[i].len / vv){
dis[to][vv] = dis[u][v] + (double)edge[i].len / vv;
pre[to][vv].u = u;
pre[to][vv].v = v;
if(!inq[to][vv]){
q.push((Node) {to , vv});
inq[to][vv] = 1;
}
}
}
}
}
inline void print(int u,int v){
if(pre[u][v].u != -1)
print(pre[u][v].u , pre[u][v].v);
printf("%d ",u);
}
int main(){
n = read(); m = read(); d = read();
for(int i=1;i<=m;i++){
a = read(); b = read(); v = read(); l = read();
add(a , b , v , l);
}
spfa();
double minn = 1e9;
int s = 0;
for(int i=0;i<=500;i++)
if(dis[d][i] < minn){
minn = min(minn , dis[d][i]);
s = i;
}
print(d , s);
return 0;
}