CSU 1806 Toll（数学+最短路）

最新推荐文章于 2017-08-20 14:29:57 发布

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图论-最短路

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本文介绍了一种结合辛普森法则与最短路径算法来解决随时间变化的路径费用问题的方法。具体而言，对于一个有向图，在特定的时间段内，每条边的费用会随着时间线性增长，目标是最小化从起点到终点的平均费用。通过使用Dijkstra算法找到不同时间点的最短路径，并利用辛普森法则进行数值积分，最终得到整个时间段内的平均最小费用。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

思路：学习了一波辛普森公式求积分点击打开链接，然后就是套最短路模板就可以了....

坑点：是有向图，看成了无向图WA了好多次

#include<bits/stdc++.h>
using namespace std;
#define inf 1e9
#define eps 1e-8
int n,m,T;
struct Node
{
	int v,c,d;
	Node(){};
	Node(int vv,int cc,int dd):v(vv),c(cc),d(dd){};
};
vector<Node>e[15];
double dis[15];
struct node
{
	int x;
	double val;
	node(){};
	node(double a,int b):val(a),x(b){};
	bool operator <(const node&rhs)const
	{
		return val>rhs.val;
	}
};
double dijkstra(int s,double t)
{
	for(int i = 1;i<=n;i++)dis[i]=inf;
	dis[s]=0;
	priority_queue<node>q;
	q.push(node(0,1));
	while(!q.empty())
	{
		node a = q.top();q.pop();
		int u = a.x;
		for(int i = 0;i<e[u].size();i++)
		{
			int v = e[u][i].v;
			double tmp = 1.0*e[u][i].c*t+e[u][i].d;
			if(dis[v]>tmp+dis[u])
			{
				dis[v]=tmp+dis[u];
				q.push(node(dis[v],v));
			}
		}
	}
	return dis[n];
}
double gaogao(double a,double b)
{
	double fa = dijkstra(1,a);
	double fb = dijkstra(1,b);
	double fc = dijkstra(1,(a+b)/2);
	return (b-a)*(fa+fb+4*fc)/6;
}
double gao(double a,double b)
{
	double mid = (a+b)/2;
	double s0 = gaogao(a,b);
	double s1 = gaogao(a,mid);
	double s2 = gaogao(mid,b);
	if(fabs(s0-s1-s2)<eps)return s0;
	else
		return gao(a,mid)+gao(mid,b);
}
int main()
{
    while(scanf("%d%d%d",&n,&m,&T)!=EOF)
	{
		for(int i = 0;i<=n;i++)
			e[i].clear();
        for(int i =1 ;i<=m;i++)
		{
			int u,v,cc,dd;
			scanf("%d%d%d%d",&u,&v,&cc,&dd);
			e[u].push_back(Node(v,cc,dd));
		}
		printf("%.8lf\n",gao(0,1.0*T)/T);
	}
}

Description

In ICPCCamp, there are n cities and m unidirectional roads between cities. The i-th road goes from the a _i-th city to the b _i-th city. For each pair of cities u and v, there is at most one road from u to v.

As traffic in ICPCCamp is becoming heavier, toll of the roads also varies. At time t, one should pay (c _i⋅t+d _i) dollars to travel along the i-th road.

Bobo living in the 1-st city would like to go to the n-th city. He wants to know the average money he must spend at least if he starts from city 1 at t∈[0,T]. Note that since Bobo's car is super-fast, traveling on the roads costs him no time.

Formally, if f(t) is the minimum money he should pay from city 1 to city n at time t, Bobo would like to find

Input

The first line contains 3 integers n,m,T (2≤n≤10,1≤m≤n(n-1),1≤T≤10 ⁴).

The i-th of the following m lines contains 4 integers a _i,b _i,c _i,d _i (1≤a _i,b _i≤n,a _i≠b _i,0≤c _i,d _i≤10 ³).

It is guaranteed that Bobo is able to drive from city 1 to city n.

Output

A floating number denotes the answer. It will be considered correct if its absolute or relative error does not exceed 10 ^-6.

Sample Input

Sample Output

1.75000000
2.00000000