poj2457 - Part Acquisition (最短路径问题)(邻接表 Dijkstra + 并查集 )

Part Acquisition

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3709 Accepted: 1599 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.

Source

USACO 2005 February Silver

 

/********************************acm:   poj-2457**title: Part Acquisition**time : 2014.8.28********************************//*    题意：          奶牛身上有物品1，现在他们要换成物品k，现在给出一张有向图，i指向j表示可以用i换j，          问最少要几次才可以换到需要的物品；如果换不到，输出-1。     输入输出：          第一行输入N(可换的物品条目 即路径总数)和K(物品数量)          接下来N行表示每行2个变量，第一个变量（物品）可以换到第二个变量（物品）。          如果换得到则输出次数，并输出路径，否则-1.    思路：          求1->k的最短路径，并记录最短路径。 本题用Dijkstra算法          第一个物品到第二个物品路径设为1，否则为无穷大*//*    测试用例  1 2 1 2 => 2 1 2              1 3 2 3 => -1*/#include <stdio.h>#include <stdlib.h>#define MAXVEX 1001#define MAXEDGE 50001#define INFINITY 100000#define TRUE 1#define FALSE 0typedef int Status;typedef struct EdgeNode{    int adjvex;    int weight;    struct EdgeNode *next;} EdgeNode;typedef struct VertexNode{//   int data;    EdgeNode *firstedge;}VertexNode, AdjList[MAXVEX];typedef struct graphAdjList{    AdjList adjList;    int numVertexes;    int numEdges;   //图中当前顶点数和边数} graphAdjList, *GraphAdjList;//创建邻接表void CreateALGraph(GraphAdjList GL, int n, int k){    int i, j;    int t;    EdgeNode *e;    GL->numVertexes = k;    GL->numEdges = n;    for (i = 1; i <= GL->numVertexes; i++)    {        GL->adjList[i].firstedge = NULL;    }    for (t = 1; t <= GL->numEdges; t++)    {        scanf("%d%d", &i, &j);        e = (EdgeNode *)malloc(sizeof(EdgeNode));        e->adjvex = j;        e->weight = 1;        e->next = GL->adjList[i].firstedge;        GL->adjList[i].firstedge = e;    } /* 静态数据    GL->numVertexes = 5;    GL->numEdges = 6;    e = (EdgeNode *)malloc(sizeof(EdgeNode));    e->adjvex = 3;    e->weight = 1;    e->next = GL->adjList[1].firstedge;    GL->adjList[1].firstedge = e;    e = (EdgeNode *)malloc(sizeof(EdgeNode));    e->adjvex = 2;    e->weight = 1;    e->next = GL->adjList[3].firstedge;    GL->adjList[3].firstedge = e;    e = (EdgeNode *)malloc(sizeof(EdgeNode));    e->adjvex = 3;    e->weight = 1;    e->next = GL->adjList[2].firstedge;    GL->adjList[2].firstedge = e;    e = (EdgeNode *)malloc(sizeof(EdgeNode));    e->adjvex = 1;    e->weight = 1;    e->next = GL->adjList[3].firstedge;    GL->adjList[3].firstedge = e;    e = (EdgeNode *)malloc(sizeof(EdgeNode));    e->adjvex = 5;    e->weight = 1;    e->next = GL->adjList[2].firstedge;    GL->adjList[2].firstedge = e;    e = (EdgeNode *)malloc(sizeof(EdgeNode));    e->adjvex = 4;    e->weight = 1;    e->next = GL->adjList[5].firstedge;    GL->adjList[5].firstedge = e; /*   int i;    for (i = 1; i <= 5; i++)      printf("%d  ", P[i]);        -1 3 -1 5 2        1  2 3  4 5*/}typedef int Patharc[MAXVEX];typedef int ShortPathTable[MAXVEX];//邻接表  Dijkstra算法  + 并查集Status ShortestPath_Dijkstra(graphAdjList GL, int v0, Patharc P){    int uniun_Eq(ShortPathTable D, int start, int end);    int v, w, k, min;    int final[MAXVEX] = {0};    ShortPathTable D;    EdgeNode *e;    for (v = 1; v <= GL.numVertexes; v++)    {        D[v] = INFINITY;        P[v] = -1;    }    e = GL.adjList[v0].firstedge;    if (e == NULL)    {        return FALSE;    }    while (e != NULL)    {        D[e->adjvex] = e->weight;        P[e->adjvex] = v0;        e = e->next;    }    final[v0] = 1;    for (v = 2; v <= GL.numVertexes; v++)    {        min = INFINITY;        for (w = 1; w <= GL.numVertexes; w++)        {            if (!final[w] && D[w] < min)            {                k = w;                min = D[w];            }        }        final[k] = 1;        e = GL.adjList[k].firstedge;        while (e != NULL)        {            if (!final[e->adjvex] && min + e->weight < D[e->adjvex])            {                D[e->adjvex] = min + e->weight;                P[e->adjvex] = k;            }            e = e->next;        }    }    return uniun_Eq(P, 1, GL.numVertexes);}//并查集 判断1与K是否构成回路int find_Eq(Patharc P, int i){  //  printf("%d\n", P[i]);    while(P[i] > -1)    {        i = P[i];    }    return i;}int uniun_Eq(Patharc P, int start, int end){    int fi;    int fj;    fi = find_Eq(P, start);    fj = find_Eq(P, end);    if (fi != fj)  //表明是两颗子树    {        return FALSE;    }    return TRUE;  //可构成回路}int main(){    int N, K;    int i;    graphAdjList GL;    Patharc P;    int p[MAXVEX];    int num;    int flag;    while (~scanf("%d%d", &N, &K))    {        if (N == 0 && K == 1)  // 特殊没有路径 只有1个顶点        {            printf("0\n1\n");            continue;        }        CreateALGraph(&GL, N, K);        flag = ShortestPath_Dijkstra(GL, 1, P);  //构成回路 TRUE,否则 FLASE        num = 0;        i = K;        if (flag == TRUE)   //判断 如果构成回路 即有一条通往终点的路径        {            do            {                p[num++] = P[i];                i = P[i];            }            while (P[i] != -1);            printf("%d\n", num+1);            for (num--; num >= 0; num--)            {                printf("%d\n", p[num]);  //输出路径            }            printf("%d\n", K);        }        else        {            printf("-1\n");        }    }    return 0;}