poj 2457

Part Acquisition

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4693 Accepted: 1998 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 51 33 22 33 12 55 4

Sample Output

41325

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.

题解：牛从第一件物品，通过交易来换取第k件物品，问最短的交换方式，典型的dijkstra的题，只不过多加了存储路径，

用path数组来存储路径，每次记录节点的后继，最后用path2来存储整条路径，最后输出。

代码：

#include<stdio.h>#include<stdlib.h>#include<math.h>#include<string.h>#define INF 0x3f3f3fint map[1007][1007], path[1007], distance[1007], visited[1007], path2[1007];int n, k;void dijkstra() {int i, j, min = INF, flag;for(i = 1; i <= k; i++)distance[i] = map[1][i];visited[1] = 1;for(i = 1; i < k; i++) {min = INF;for(j = 1; j <= k; j++)if(min > distance[j] && !visited[j]) {min = distance[j];flag = j;}visited[flag] = 1;for(j = 1; j <= k; j++)if(distance[j] > min + map[flag][j] && !visited[j]) {distance[j] = min + map[flag][j];path[j] = flag;}}}void outPut() {int i, count = 0, temp;memset(path2, 0, sizeof(path2));if(distance[k] == INF)printf("-1\n");else {temp = path[k];path2[count++] = k;while(temp != -1) {path2[count++] = temp;temp = path[temp];}printf("%d\n", count + 1);printf("1\n");for(i = count - 1; i >= 0; i--)printf("%d\n", path2[i]);}}int main() {int i, j, start, end;while(scanf("%d%d", &n, &k) != EOF) {for(i = 1; i < 1007; i++)for(j = 1; j < 1007; j++)map[i][j] = INF;memset(visited, 0, sizeof(visited));memset(path, -1, sizeof(path));for(i = 1; i <= n; i++) {scanf("%d%d", &start, &end);map[start][end] = 1;}dijkstra();outPut();}}