PAT 甲级 1030  Travel Plan

本文介绍了一种解决旅行者地图上从起点到目的地最短路径问题的算法,该算法考虑了城市间的距离和成本,确保找到的路径不仅是最短的,而且是成本最低的。输入包括城市数量、高速公路数量、起点和终点,以及每条高速公路的距离和成本。通过Dijkstra算法实现,输出最短路径上的城市序列、总距离和总成本。

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1030 Travel Plan (30 point(s))

A traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her starting city and the destination. If such a shortest path is not unique, you are supposed to output the one with the minimum cost, which is guaranteed to be unique.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 4 positive integers N, M, S, and D, where N (≤500) is the number of cities (and hence the cities are numbered from 0 to N−1); M is the number of highways; S and D are the starting and the destination cities, respectively. Then M lines follow, each provides the information of a highway, in the format:

City1 City2 Distance Cost

where the numbers are all integers no more than 500, and are separated by a space.

Output Specification:

For each test case, print in one line the cities along the shortest path from the starting point to the destination, followed by the total distance and the total cost of the path. The numbers must be separated by a space and there must be no extra space at the end of output.

Sample Input:

4 5 0 3
0 1 1 20
1 3 2 30
0 3 4 10
0 2 2 20
2 3 1 20

Sample Output:

0 2 3 3 40

Experiential Summing-up

This question is classical short distance problem. Just have two edge weight in the progress of comparing.That's all~

(The purpose of using English to portray my solution is that to exercise the ability of my expression of English and accommodate PAT advanced level's style.We can make progress together by reading and comprehending it. Please forgive my basic grammar's and word's error. Of course, I would appreciated it if you can point out my grammar's and word's error in comment section.( •̀∀•́ ) Furthermore, Big Lao please don't laugh at me because I just a English beginner settle for CET6    _(:з」∠)_  )

Accepted Code

#include <cstdio>
#include <algorithm>
#include <vector>
#include <cstring>
using namespace std;
const int maxn=510;
const int INF=0x3fffffff;

vector<int> path;
int n,m,s,D,d[maxn],c[maxn],pre[maxn];
bool vis[maxn];
struct node
{
	int v,w,c;
	node(int a,int b,int d):v(a),w(b),c(d){};
};
vector<node> adj[maxn];
void Dijkstra()
{
	fill(d,d+maxn,INF);
	fill(c,c+maxn,INF);
	memset(vis,false,sizeof(vis));
	for(int i=0;i<n;++i)
		pre[i]=i;
	d[s]=0;
	c[s]=0;
	for(int i=0;i<n;++i)
	{
		int u=-1,min=INF;
		for(int j=0;j<n;++j)
		{
			if(vis[j]==false&&d[j]<min)
			{
				min=d[j];
				u=j;
			}
		}
		if(u==-1)
			return ;
		vis[u]=true;
		for(int j=0;j<adj[u].size();++j)
		{
			int v=adj[u][j].v;
			if(vis[v]==false)
			{
				int w=adj[u][j].w;
				int x=adj[u][j].c;
				if(d[u]+w<d[v])
				{
					d[v]=d[u]+w;
					c[v]=c[u]+x;
					pre[v]=u;
				}
				else if(d[u]+w==d[v]&&c[u]+x<c[v])
				{
					c[v]=c[u]+x;
					pre[v]=u;
				}
			}
		}
	}
}
void DFS(int x)
{
	if(x==s)
	{
		path.push_back(x);
		return;
	}
	path.push_back(x);
	DFS(pre[x]);
}
int main()
{
	int e1,e2,dis,cost;
	scanf("%d %d %d %d",&n,&m,&s,&D);
	for(int i=0;i<m;++i)
	{
		scanf("%d %d %d %d",&e1,&e2,&dis,&cost);
		adj[e1].push_back(node(e2,dis,cost));
		adj[e2].push_back(node(e1,dis,cost));
	}
	Dijkstra();
	DFS(D);
	for(int i=path.size()-1;i>=0;--i)
	{
		printf("%d ",path[i]);
	}
	printf("%d %d\n",d[D],c[D]);
	return 0;
}

 

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