洛谷P1266速度限制

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;
}