狂飆突進的幻想鄉
阿新 • • 發佈:2018-12-10
題目大意:給一張無向圖,每條邊有個(x,y),每條邊的邊權是。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));
}