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
;
vis =new int
;
dis =new int
;
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