PAT Advanced Level 1030. Travel Plan (30)(Java and C++)

PAT Advanced Level 1030. Travel Plan (30)

最短路径问题

我的思路：

1、SPFA算法，求出各个点的最短路径

2、DFS遍历，并将个各条到达目的地且路径最短的路线存储到pathVC

3、遍历pathVC求出minCost

看到别的大神有更加简便的思路：

思路（使用Dijkstra或者SPFA）：比较大小时，将距离相同但花费更短也作为更新的条件之一

            //比较大小时，将距离相同但花费更短也作为更新的条件之一              if (d[t]>d[v] + e.dis || (d[t] == d[v] + e.dis&&cost[t] > cost[v] + e.cost))              {                  d[t] = d[v] + e.dis;                  cost[t] = cost[v] + e.cost;                  pre[t] = v;                  que.push(P(d[t], t));              }

Java代码如下（AC）：

import java.util.LinkedList;import java.util.Queue;import java.util.Scanner;import java.util.Vector;public class Main{static Main  instance =new Main();static int N;static int M;static int S;static int D;static int[] dis;static int[] cost;static int[] vis;static Vector<Edge>[] map ;static final  int Inf  =Integer.MAX_VALUE-10000;public static void main(String[] args){Scanner sc =new Scanner(System.in);N =sc.nextInt();M =sc.nextInt();S =sc.nextInt();D =sc.nextInt();map =new Vector[N];vis =new int[N];dis =new int[N];for(int i =0;i<N;i++){map[i]=new Vector<Edge>();vis[i]=0;dis[i]=Inf;}cost=new int[M];for(int i=0;i<M;i++){int  c1 = sc.nextInt();int  c2 = sc.nextInt();int  distance = sc.nextInt();cost[i] = sc.nextInt();Edge e1 =instance.new Edge();e1.to=c2;e1.len  = distance;e1.eCost = cost[i];map[c1].add(e1);Edge e2 =instance.new Edge();e2.to=c1;e2.len =  distance;e2.eCost= cost[i];map[c2].add(e2);}SPFA(S);DFS(S,D,0);path.clear();int idx =-1;int minCost=Integer.MAX_VALUE;int[] shortedCost =new int[pathVC.size()];for(int i=0;i<pathVC.size();i++){path=pathVC.get(i);for(int j=0;j<path.size();j++){shortedCost[i] +=path.get(j).eCost;}if(minCost > shortedCost[i]){   minCost = shortedCost[i];   idx     = i;}}path =pathVC.get(idx);System.out.print(S+" ");for(int i=0;i<path.size();i++){System.out.print(path.get(i).to+" ");}System.out.print(dis[D]);System.out.print(" ");System.out.print(minCost);}public static void SPFA(int st){Queue<Integer> q =new LinkedList<Integer>();q.add(st);dis[st]=0;vis[st]=1;while(!q.isEmpty()){int cur =q.poll();Vector<Edge> adjEdges =map[cur];for(int i =0;i<adjEdges.size();i++){int to  =adjEdges.get(i).to;int len =adjEdges.get(i).len;                if(dis[cur]+len <dis[to]){                   dis[to]=dis[cur]+len;                if(vis[to]==0){                vis[to]=1;                q.add(to);                }                }}}}static Vector<Vector<Edge>>  pathVC =new Vector<Vector<Edge>>();static Vector<Edge>  path =new Vector<Edge>();public static void DFS(int st ,int des ,int distance){if(st==des){if(distance==dis[des]){Vector<Edge>  onePath =new Vector<Edge>();for(int i=0;i<path.size();i++){onePath.add(path.get(i));}pathVC.add(onePath);}path.clear();return;}if(distance>dis[st]){path.clear();return;}Vector<Edge> adjEdges =map[st];for(int i=0;i<adjEdges.size();i++){Edge e =adjEdges.get(i);int to  =e.to;int len =e.len;if(dis[st]+len<=dis[to]){path.add(e);DFS(to,des,dis[to]);}}path.clear();}class Edge{int to;int len;int eCost;}}

参考博客：

http://blog.csdn.net/gemire/article/details/20698375