題目大意：給一張無向圖，每條邊有個(x,y)，每條邊的邊權是ax+(1−a)y$ax+(1-a)y$。若a在[0,1]均勻隨機，問期望最短路是多少。n<=200,m<=400。x,y<=10^7且隨機生成。 題解：你可以直接辛普森積分水過。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<climits>
#include<cstdlib>
#include<ctime>
 

#include<queue>
#include<vector>
#include<utility>
#define gc getchar()
#define lint long long
#define db long double
#define pb push_back
#define mp make_pair
#define fir first
#define sec second
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define Rep(i,v) rep(i,0,(int)v.size()-1)
 

#define INF 2.0*INT_MAX
#define N 210
#define M 410
#define debug(x) cerr<<#x<<"="<<x
#define sp <<" "
#define ln <<endl
using namespace std;
inline int inn()
{
    int x,ch;while((ch=gc)<'0'||ch>'9');
    x=ch^'0';while((ch=gc)>='0'&&ch<='9')
        x=(x<<1
 
)+(x<<3)+(ch^'0');return x;
}
const db eps=0.00000001;inline db gabs(db x) { return x<0?-x:x; }
priority_queue<pair<db,int> > q;db d[N],w[M];bool vis[N];
vector<pair<int,db> > g[N];int u[M],v[M],x[M],y[M],n,m,s,t;
inline db dijkstra(db a)
{
    rep(i,1,n) d[i]=INF,vis[i]=0;
    while(!q.empty()) q.pop();q.push(mp(d[s]=0,s));
    rep(i,1,n) g[i].clear();rep(i,1,m) w[i]=a*x[i]+(1-a)*y[i];
    rep(i,1,m) g[u[i]].pb(mp(v[i],w[i])),g[v[i]].pb(mp(u[i],w[i]));
    while(!q.empty())
    {
        int x=q.top().sec;q.pop();if(vis[x]) continue;
        if(x==t) return d[t];vis[x]=1;int y;
        Rep(i,g[x]) if(!vis[y=g[x][i].fir]) if(d[y]>d[x]+g[x][i].sec)
            d[y]=d[x]+g[x][i].sec,q.push(mp(-d[y],y));
    }
    return d[t];
}
db simpson(db l,db r,db vl,db vr,db vm)
{
    db mid=(l+r)/2,lm=(l+mid)/2,rm=(mid+r)/2;
    db vlm=dijkstra(lm),vrm=dijkstra(rm);
    db sl=(mid-l)*(vl+vm+4*vlm)/6,
       sr=(r-mid)*(vm+vr+4*vrm)/6,
        s=(r-l)*(vl+vr+4*vm)/6;
    if(gabs(r-l)<eps||gabs(sl+sr-s)<eps) return (sl+sr+s)/2;
    return simpson(l,mid,vl,vm,vlm)+simpson(mid,r,vm,vr,vrm);
}
db simpson(db L,db R) { return simpson(L,R,dijkstra(L),dijkstra(R),dijkstra((L+R)/2)); }
int main()
{
    n=inn(),m=inn(),s=inn(),t=inn();
    rep(i,1,m) u[i]=inn(),v[i]=inn(),x[i]=inn(),y[i]=inn();
    return !printf("%lf\n",(double)simpson(0,1));
}