URAL 1980 Road to Investor

such that the maximal needed overspeeding along this route is as minimal as possible“一般情况”，看到这句就是二分了。二分答案，跑最短路即可。路径还原什么的。。应该不是问题吧

/*
    author:ray007great
*/
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
/*  define */

#define sf(a) scanf("%d",&a)
#define sfs(a) scanf("%s",a)
#define sfI(a) scanf("%I64d",&a)
#define pf(a) printf("%d\n",a)
#define pfI(a) printf("%I64d\n",a)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repd(i,a,b) for(int i=(a);i>=(b);i--)
#define rep1(i,a,b) for(int i=(a);i<(b);i++)
#define clr(a) memset(a,0,sizeof(a))
#define pfk  printf("fuck\n")

/*  define */
queue q;
const int N = 11000;
const double inf = 9999999;
const double eps = 1e-8;
struct edge{
    int u,v;
    double l,s;
    int id;
};
vector g[N];
vector ans;
bool inq[N];
int n,m,path[N];
double d[N];
int fa[N];
double t;
void spfa(double plus){
    memset(inq,0,sizeof(inq));
    rep(i,2,n) d[i]=inf;
    d[1]=0;
    q.push(1);
    while(!q.empty()){
      //  pfk;
        int u=q.front();q.pop();
        inq[u]=false;
        int sz=g[u].size();
        rep1(i,0,sz){
            int v=g[u][i].v;
            double w=g[u][i].l/(g[u][i].s+plus);
            if(d[v]>d[u]+w){
                d[v]=d[u]+w;
                fa[v]=u;
                path[v]=g[u][i].id;
                if(!inq[v]){
                    inq[v]=true;
                    q.push(v);
                }
            }
        }
    }
}
bool ok(double x){
    spfa(x);
    if(d[n]<=t) {//
       // pfk;
    //   printf("%lf\n",d[n]);
        return true;
    }
    return false;
}
int main(){
    while(~scanf("%d%d",&n,&m)){
        ans.clear();
        rep(i,1,n) g[i].clear();
        rep(i,1,m){
            int u,v;
            double li,s;
            scanf("%d%d%lf%lf",&u,&v,&s,&li);
            edge t1,t2;
            t1.u=u;t1.v=v;t1.l=li;t1.s=s;t1.id=i;
            t2.u=u;t2.v=v;t2.l=li;t2.s=s;t2.id=i;
            g[u].push_back(t1);
            g[v].push_back(t2);
        }
        scanf("%lf",&t);
        double l=0,r=1e8,m;
        while(fabs(l-r)>eps){
           // pfk;
            m=(l+r)/2;
            if(ok(m)) r=m;
            else l=m;
         //   printf("%lf\n",m);
        }
        for(int i=n;i!=1;i=fa[i]){
            ans.push_back(path[i]);
        }
        int sz=ans.size();
        printf("%.6f %d\n",m,sz);
        printf("%d",ans[sz-1]);
        for(int i=sz-2;i>=0;i--) printf(" %d",ans[i]);
        puts("");
    }
    return 0;
}