CSU 1806 Toll 自适应simpson积分+最短路

分析：根据这个题学了一发自适应simpson积分（原来积分还可以这么求），然后就是套模板了

学习自适应simpson积分:http://blog.csdn.net/greatwall1995/article/details/8639135

#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 1e2 + ;
const double eps = 1e-;
const double INF = 1e12;
int n,m,T,tot,head[];
int a[N],b[N],c[N],d[N];
bool vis[];
double dis[N];
struct Edge{
  int v,next;
  double w;
  Edge(int v=,double w=){
    this->v=v;this->w=w;
  }
  bool operator<(const Edge &rhs)const{
    return w>rhs.w;
  }
}edge[N];
void add(int u,int v,double w){
  edge[tot].v=v;
  edge[tot].w=w;
  edge[tot].next=head[u];
  head[u]=tot++;
}
priority_queue<Edge>q;
double F(double t){
   memset(head,-,sizeof(head));tot=;
   memset(vis,false,sizeof(vis));
   for(int i=;i<=n;++i)dis[i]=INF;
   dis[]=;
   for(int i=;i<m;++i){
    add(a[i],b[i],c[i]*t+d[i]);
   }
   q.push(Edge(,dis[]));
   while(!q.empty()){
      int u=q.top().v;
      q.pop();if(vis[u])continue;
      vis[u]=true;
      for(int i=head[u];~i;i=edge[i].next){
          int v=edge[i].v;
          if(!vis[v]&&dis[v]>dis[u]+edge[i].w){
             dis[v]=dis[u]+edge[i].w;
             q.push(Edge(v,dis[v]));
          }
      }
   }
   return dis[n];
}
double simpson(double a,double b){
  double c=a+(b-a)/;
  return (F(a)+*F(c)+F(b))*(b-a)/;
}
double asr(double a,double b,double eps,double A){
    double c=a+(b-a)/;
    double L = simpson(a,c),R=simpson(c,b);
    if(fabs(L+R-A)<=*eps)return L+R+(L+R-A)/15.0;
    return asr(a,c,eps/,L)+asr(c,b,eps/,R);
}
double get(double a,double b,double eps){
  return asr(a,b,eps,simpson(a,b));
}
int main(){
  while(~scanf("%d%d%d",&n,&m,&T)){
    for(int i=;i<m;++i)scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
    printf("%.6f\n",get(,T,eps)/T);
  }
  return ;
}