[Dijkstra] Highway Project 16浙江省赛K

最新推荐文章于 2019-04-18 22:19:04 发布

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最新推荐文章于 2019-04-18 22:19:04 发布

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分类专栏：最短路

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本文介绍了一个算法问题，即如何为Marjar帝国构建最优的高速公路网络，使得从首都到达其他城市的总时间最短，并且在满足此条件的基础上成本最低。通过使用Dijkstra算法，解决了这一复杂的图论问题。

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Edward, the emperor of the Marjar Empire, wants to build some bidirectional highways so that he can reach other cities from the capital as fast as possible. Thus, he proposed the highway project.

The Marjar Empire has N cities (including the capital), indexed from 0 to N - 1 (the capital is 0) and there are M highways can be built. Building the i-th highway costsC_i dollars. It takes D_i minutes to travel between city X_i and Y_i on the i-th highway.

Edward wants to find a construction plan with minimal total time needed to reach other cities from the capital, i.e. the sum of minimal time needed to travel from the capital to city i (1 ≤ i ≤ N). Among all feasible plans, Edward wants to select the plan with minimal cost. Please help him to finish this task.

Input

There are multiple test cases. The first line of input contains an integer Tindicating the number of test cases. For each test case:

The first contains two integers N, M (1 ≤ N, M ≤ 10⁵).

Then followed by M lines, each line contains four integers X_i, Y_i, D_i, C_i (0 ≤ X_i,Y_i < N, 0 < D_i, C_i < 10⁵).

<h4< dd="">Output

For each test case, output two integers indicating the minimal total time and the minimal cost for the highway project when the total time is minimized.

<h4< dd="">Sample Input

<h4< dd="">Sample Output

4 3
4 4

#include <bits/stdc++.h>
using namespace std;
long long int d[100005];
long long int mo[100005];
//mo存到该点的最短一截路 
struct WAY
{
    int to;
    long long int cost;
    long long int fee;
};
vector<WAY>way[100005];
struct node
{
	int num;
	long long int dis;
};
bool operator <(node a,node b)
{
	return a.dis>b.dis;
}  //取距离最短的点 
priority_queue<node>que;
void dijkstra()
{
	while(!que.empty())
	{
		node now=que.top();que.pop();
		int nu=now.num;int di=now.dis;
		int L=way[nu].size();
		for(int i=0;i<L;i++)
		{
			WAY w=way[nu][i];
			if(d[w.to]>(d[nu]+w.cost))
			{
				mo[w.to]=w.fee;
				d[w.to]=d[nu]+w.cost;
				node no;
				no.num=w.to;no.dis=d[w.to];
				que.push(no);
			}
			else if(d[w.to]==(d[nu]+w.cost))
				mo[w.to]=min(mo[w.to],w.fee);
		}
	}
}
int main()
{
	int T;
	cin>>T;
	while(T--)
	{
		for(int i=0;i<100005;i++)
		    way[i].clear(); 
		int n,m;
		cin>>n>>m;
		while(m--)
		{
			int x,y,d,c;
			cin>>x>>y>>d>>c;
			WAY w;
			w.to=y;w.cost=d;w.fee=c;
			way[x].push_back(w);
			w.to=x;
			way[y].push_back(w);
		}
		for(int i=1;i<n;i++)
		   d[i]=1e18;
		d[0]=0;
		node w;w.num=0;w.dis=0;
		que.push(w);
		dijkstra();
		long long int sum=0;
		for(int i=0;i<n;i++)
		   sum+=d[i];
		long long int money=0;
		for(int i=1;i<n;i++)
		   money+=mo[i];
		cout<<sum<<' '<<money<<endl;
	}
	return 0;
 }