POJ 2457 Part Acquisition【Dij+记录路径】

Part Acquisition

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4128		Accepted: 1767		Special Judge

Description

The cows have been sent on a mission through space to acquire a new milking machine for their barn. They are flying through a cluster of stars containing N (1 <= N <= 50,000) planets, each with a trading post. 

The cows have determined which of K (1 <= K <= 1,000) types of objects (numbered 1..K) each planet in the cluster desires, and which products they have to trade. No planet has developed currency, so they work under the barter system: all trades consist of each
party trading exactly one object (presumably of different types). 

The cows start from Earth with a canister of high quality hay (item 1), and they desire a new milking machine (item K). Help them find the best way to make a series of trades at the planets in the cluster to get item K. If this task is impossible, output -1.

Input

* Line 1: Two space-separated integers, N and K. 

* Lines 2..N+1: Line i+1 contains two space-separated integers, a_i and b_i respectively, that are planet i's trading trading products. The planet will give item b_i in order to receive item a_i.

Output

* Line 1: One more than the minimum number of trades to get the milking machine which is item K (or -1 if the cows cannot obtain item K). 

* Lines 2..T+1: The ordered list of the objects that the cows possess in the sequence of trades.

Sample Input

6 5

1 3

3 2

2 3

3 1

2 5

5 4

Sample Output

4

1

3

2

5

Hint

OUTPUT DETAILS: 

The cows possess 4 objects in total: first they trade object 1 for object 3, then object 3 for object 2, then object 2 for object 5.

Source

USACO 2005 February Silver

 

题目大意：有n条边，K个点，我们的目的是从1作为源点，K作为终点，求一条从1-k的最短路，并输出路径，如果没有边能从1走到K，那么输出-1.

思路：

1、N比较大，K 反而比较小，我们采用dijkstra算法来求最短路，1000*1000对于1000ms来说微不足道。

2、记录路径问题，我们设定path【i】=x，表示点i作为终点的最短路径从x点走过来，也就是说，在松弛的时候：dis【v】=dis【u】+map【u】【v】的时候，我们设定path【v】=u，每次都随着更新，path【v】里边的u一定是最后一次松弛的点。这样，我们在初始的时候，将path【1-k】都设定为值-1，然后我们一直使得now=path【now】，然后记录now即可。

AC代码：
#include<stdio.h>
#include<string.h>
using namespace std;
int head[1000000];
struct EdgeNode
{
int from;
int to;
int next;
int w;
}e[1000000];
int vis[1000000];
int dis[1000000];
int path[1000000];
int ans[1000000];
int n,m,cont;
void add(int from,int to,int w)
{
e[cont].from=from;
e[cont].w=w;
e[cont].to=to;
e[cont].next=head[from];
head[from]=cont++;
}
void Dij()
{
memset(vis,0,sizeof(vis));
for(int i=1;i<=n;i++)path[i]=-1;
for(int i=1;i<=n;i++)dis[i]=0x3f3f3f3f;
dis[1]=0;
for(int i=0;i<n;i++)
{
int u=-1;
int tmp=0x3f3f3f3f;
for(int j=1;j<=n;j++)
{
if(vis[j]==0&&tmp>dis[j])
{
u=j;
tmp=dis[j];
}
}
vis[u]=1;
for(int k=head[u];k!=-1;k=e[k].next)
{
int v=e[k].to;
int w=e[k].w;
if(dis[v]>dis[u]+w)
{
dis[v]=dis[u]+w;
path[v]=u;
}
}
}
if(dis
==0x3f3f3f3f)
{
printf("-1\n");
return ;
}
int tot=0;
ans[tot++]=n;
int now=n;
while(1)
{
now=path[now];
if(now==-1)break;
ans[tot++]=now;
}
printf("%d\n",tot);
for(int i=tot-1;i>=0;i--)
{
printf("%d\n",ans[i]);
}
}
int main()
{
while(~scanf("%d%d",&m,&n))
{
cont=0;
memset(head,-1,sizeof(head));
for(int i=0;i<m;i++)
{
int x,y;
scanf("%d%d",&x,&y);
add(x,y,1);
}
Dij();
}
}