POJ 2457 Part Acquisition (Dijkstra + 记录路径)

Part Acquisition

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3767		Accepted: 1631		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

题目链接：http://poj.org/problem?id=2457

题目大意：有向图，一共有n条边，求从点1开始到点k的最短路径，不存在输出－1

题目分析：裸的Dijkstra，逆向记录下路径就好了，水题

#include <cstdio>
#include <cstring>
int const INF = 0xfffffff;
int const MAX = 1005;
int dist[MAX], map[MAX][MAX], path[MAX], ans[MAX];
int n, k;
bool vis[MAX];

void Dijkstra(int v0)
{  
    for(int i = 1; i <= k; i++)  
    {  
        dist[i] = map[v0][i];  
        vis[i] = false;  
        if(i != v0 && dist[i] < INF)
            path[i] = v0; 
    }   
    dist[v0] = 0; 
    vis[v0] = true;
    int u;
    for(int i = 1; i <= k; i++)  
    {  
        int Min = INF;  
        for(int j = 1; j <= k; j++)  
        {  
            if(!vis[j] && dist[j] < Min)  
            {  
                u = j;  
                Min = dist[j];  
            }  
        }  
        vis[u] = true;  
        for(int j = 1; j <= k; j++)  
        {
            if(!vis[j] &&  map[u][j] < INF && dist[u] + map[u][j] < dist[j]) 
            {
                dist[j] = dist[u] + map[u][j];  
                path[j] = u;
            }
        }
    }  
}  

void Print()
{
    memset(ans , 0, sizeof(ans));
    int cnt = 0;
    if(dist[k] == INF)
        printf("-1\n");
    else
    {
        ans[cnt++] = k;
        int tmp = path[k];
        while(tmp)
        {
            ans[cnt++] = tmp;
            tmp = path[tmp];
        }
        printf("%d\n", cnt);
        for(int i = cnt - 1; i >= 0; i--)
            printf("%d\n", ans[i]);
    }
}

int main()
{
    int x, y;
    scanf("%d %d", &n, &k);
    memset(path , 0, sizeof(path));
    for(int i = 1; i <= k; i++)
    {
        for(int j = 1; j <= k; j++)
        {
            map[i][j] = INF;
            if(i == j)
                map[i][j] = 0;
        }
    }
    for(int i = 0; i < n; i++)
    {
        scanf("%d %d", &x, &y);
        map[x][y] = 1;
    }
    Dijkstra(1);
    Print();
}