BZOJ3245: 最快路线 拆点dijkstra

150个点，500种速度，乘起来大概8e4个点，3e4的边

其他题解写的基本都是spfa，想想dij也能做，还挺快

#include<bits/stdc++.h>  
//#pragma comment(linker, "/STACK:1024000000,1024000000")   
#include<stdio.h>  
#include<algorithm>  
#include<queue>  
#include<string.h>  
#include<iostream>  
#include<math.h>                    
#include<set>  
#include<map>  
#include<vector>  
#include<iomanip> 
#include<bitset>
using namespace std;         //

#define ll long long  
#define pb push_back  
#define FOR(a) for(int i=1;i<=a;i++) 
#define sqr(a) (a)*(a)
#define dis(a,b) sqrt(sqr(a.x-b.x)+sqr(a.y-b.y))
ll qp(ll a,ll b,ll mod){
	ll t=1;while(b){if(b&1)t=t*a%mod;b>>=1;a=a*a%mod;}return t;
}
struct DOT{int x;int y;};
const int dx[4]={0,0,-1,1};
const int dy[4]={1,-1,0,0};
const int inf=0x3f3f3f3f;  
const ll mod=1e9+7;
const int maxn=2e2+5;       //---------------maxn---------------//

/******************************************************/
namespace fastIO{  
    #define BUF_SIZE 100000  
    #define OUT_SIZE 100000  
    #define ll long long  
    //fread->read  
    bool IOerror=0;  
    inline char nc(){  
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;  
        if (p1==pend){  
            p1=buf; pend=buf+fread(buf,1,BUF_SIZE,stdin);  
        if (pend==p1){IOerror=1;return -1;}  
        //{printf("IO error!
");system("pause");for (;;);exit(0);}  
    }  
    return *p1++;  
}  
inline bool blank(char ch){return ch==' '||ch=='
'||ch=='
'||ch=='	';}  
inline void read(int &x){  
    bool sign=0; char ch=nc(); x=0;  
    for (;blank(ch);ch=nc());  
    if (IOerror)return;  
    if (ch=='-')sign=1,ch=nc();  
    for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';  
    if (sign)x=-x;  
}  
inline void read(ll &x){  
    bool sign=0; char ch=nc(); x=0;  
    for (;blank(ch);ch=nc());  
    if (IOerror)return;  
    if (ch=='-')sign=1,ch=nc();  
    for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';  
    if (sign)x=-x;  
}  
inline void read(double &x){  
    bool sign=0; char ch=nc(); x=0;  
    for (;blank(ch);ch=nc());  
    if (IOerror)return;  
    if (ch=='-')sign=1,ch=nc();  
    for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';  
    if (ch=='.'){  
        double tmp=1; ch=nc();  
        for (;ch>='0'&&ch<='9';ch=nc())tmp/=10.0,x+=tmp*(ch-'0');  
    }  
    if (sign)x=-x;  
}  
inline void read(char *s){  
    char ch=nc();  
    for (;blank(ch);ch=nc());  
    if (IOerror)return;  
    for (;!blank(ch)&&!IOerror;ch=nc())*s++=ch;  
    *s=0;  
}  
inline void read(char &c){  
    for (c=nc();blank(c);c=nc());  
    if (IOerror){c=-1;return;}  
} 
#undef OUT_SIZE  
#undef BUF_SIZE  
}; using namespace fastIO;
/*****************************************************/


//n 150,v 500,l 500

int n,m,D;

double d[maxn][550];	//到i点速度为j的最小时间
pair<int,int> from[maxn][550];	//来源

int head[maxn],tot;
struct EDGE{
	int a,b,v,l;
	int nxt;
}edges[maxn*maxn];
void addedge(int a,int b,int v,int l){
	edges[++tot]=(EDGE){a,b,v,l,head[a]};
	head[a]=tot;
}

struct sed{
	int u;int v;double t;
	friend bool operator <(sed a,sed b){return a.t>b.t;}};
priority_queue<sed>Q;

bool vis[maxn][550];
void dij(){
	memset(d,127,sizeof d);
	Q.push((sed){0,70,0});
	d[0][70]=0;
	
	while(!Q.empty()){
		sed now=Q.top();Q.pop();
		int x=now.u,vx=now.v;double t=now.t;
		if(vis[x][vx])continue;
		vis[x][vx]=1;

		for(int i=head[x];i;i=edges[i].nxt){
			int y=edges[i].b,vy=edges[i].v,l=edges[i].l;
			if(vy && d[x][vx]+(double)l/vy < d[y][vy]){
				d[y][vy]=d[x][vx]+(double)l/vy;
				from[y][vy]=make_pair(x,vx);
				Q.push((sed){y,vy,d[y][vy]});
			}
			if(!vy && d[x][vx]+(double)l/vx<d[y][vx]){
				d[y][vx]=d[x][vx]+(double)l/vx;
				from[y][vx]=make_pair(x,vx);
				Q.push((sed){y,vx,d[y][vx]});
			}
		}
	}
}

void print(int x,int vx){
	if(x)print(from[x][vx].first,from[x][vx].second);
	printf("%d",x);
	if(x!=D)printf(" ");
}

int main(){
	read(n);read(m);read(D);
	for(int i=1;i<=m;i++){
		int a,b,v,l;
		read(a);read(b);read(v);read(l);
		addedge(a,b,v,l);
	}
	dij();
	int ans=0;
	for(int i=1;i<=500;i++){
		if(d[D][i]<d[D][ans])ans=i;
	}
	print(D,ans);
	return 0;
}