1030 Travel Plan-PAT甲级

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本文介绍了一种基于Dijkstra算法的最短路径寻找方法，旨在帮助旅行者找到从起点到终点的最短路径，当存在多条最短路径时，算法会输出成本最低的那一条。通过不断更新最短路径和最小成本，最终使用DFS输出完整路径。

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:

Sample Output:

0 2 3 3 40

求起点到终点的最短路径，如果路径有多条，则输出花费的代价最小的路径

dijkstra最短路的变形，在求解过程中，不断地更换最短路径以及最少花费即可，最后用dfs输出路径即可

满分代码如下：

#include<bits/stdc++.h>
using namespace std;
const int N=505,inf=0x3f3f3f3f;
struct Edge{
	int v,dst,cost;
	Edge(){}
	Edge(int vv,int dd,int cc){
		v=vv;
		dst=dd;
		cost=cc;
	}
};
vector<Edge>graph[N];
int vst[N],dst[N],cost[N],pre[N];
int n,m,s,d;
void dijkstra(int ss,int dd){
	memset(vst,0,sizeof(vst));
	fill(dst,dst+N,inf);
	fill(cost,cost+N,inf);
	dst[ss]=0;
	cost[ss]=0;
	while(!vst[dd]){
		//还未访问到终点;
		int u=-1,min_dist=0x3f3f3f3f;
		for(int i=0;i<n;i++){
			if(!vst[i]&&dst[i]<min_dist){
				u=i;
				min_dist=dst[i];
			}
		} 
		if(u==-1) return;
		vst[u]=1;
		for(auto i:graph[u]){
			if(dst[u]+i.dst<dst[i.v]){
				dst[i.v]=dst[u]+i.dst;
				pre[i.v]=u;
				cost[i.v]=cost[u]+i.cost;
			}else if(dst[u]+i.dst==dst[i.v]&&cost[u]+i.cost<cost[i.v]){
				pre[i.v]=u;
				cost[i.v]=cost[u]+i.cost;
			}
		}
	}
}
int flag=0;
void dfs(int s,int d){
	if(d==s){
		printf("%d",s);
		return;
	}
	dfs(s,pre[d]);
	printf(" %d",d);
}
int main(){
	scanf("%d%d%d%d",&n,&m,&s,&d);
	for(int i=1;i<=m;i++){
		int u,v,dst,cost;
		scanf("%d%d%d%d",&u,&v,&dst,&cost);
		graph[u].push_back({v,dst,cost});
		graph[v].push_back({u,dst,cost});
	}
	dijkstra(s,d);
	dfs(s,d);
	printf(" %d %d\n",dst[d],cost[d]);
	return 0; 
}