4-1 Shortest Path [3]

4-1 Shortest Path [3]   (10分)

Write a program to not only find the weighted shortest distances, but also count the number of different minimum paths from any vertex to a given source vertex in a digraph. It is guaranteed that all the weights are positive.

Format of functions:

void ShortestDist( MGraph Graph, int dist[], int count[], Vertex S );

where MGraph is defined as the following:

typedef struct GNode *PtrToGNode;struct GNode{    int Nv;    int Ne;    WeightType G[MaxVertexNum][MaxVertexNum];};typedef PtrToGNode MGraph;

The shortest distance from V to the source S is supposed to be stored in dist[V]. If V cannot be reached from S, store -1 instead. The number of different minimum paths from V to the source S is supposed to be stored in count[V] and count[S]=1.

Sample program of judge:

#include <stdio.h>#include <stdlib.h>typedef enum {false, true} bool;#define INFINITY 1000000#define MaxVertexNum 10  /* maximum number of vertices */typedef int Vertex;      /* vertices are numbered from 0 to MaxVertexNum-1 */typedef int WeightType;typedef struct GNode *PtrToGNode;struct GNode{    int Nv;    int Ne;    WeightType G[MaxVertexNum][MaxVertexNum];};typedef PtrToGNode MGraph;MGraph ReadG(); /* details omitted */void ShortestDist( MGraph Graph, int dist[], int count[], Vertex S );int main(){    int dist[MaxVertexNum], count[MaxVertexNum];    Vertex S, V;    MGraph G = ReadG();    scanf("%d", &S);    ShortestDist( G, dist, count, S );    for ( V=0; V<G->Nv; V++ )        printf("%d ", dist[V]);    printf("\n");    for ( V=0; V<G->Nv; V++ )        printf("%d ", count[V]);    printf("\n");    return 0;}/* Your function will be put here */

Sample Input (for the graph shown in the figure):

8 110 4 50 7 101 7 303 0 403 1 203 2 1003 7 704 7 56 2 17 5 37 2 503

Sample Output:

40 20 100 0 45 53 -1 50 
1 1 4 1 1 3 0 3
void ShortestDist( MGraph Graph, int dist[], int count[], Vertex S )  {  int visit[MaxVertexNum];  int dist1[MaxVertexNum];  int i;       /*初始化*/  for(i=0;i<Graph->Nv;i++){  dist[i]=Graph->G[S][i];   /*dist[i]代表从S到i的最短距离 */         count[i]=0; /* count[i]表示从S到i的最短路径条数 */          visit[i]=0; /* 未访问过该结点 */  }  //原点的初始化visit[S]=1;  dist[S]=0;     /*书上的dijikstra算法的实现*/   while(1){  int min=INFINITY;  int v=-1;     for(i=0;i<Graph->Nv;i++){     //遍历节点，找到距离最小的未被访问的结点if(!visit[i]&&dist[i]<min){  min=dist[i];  v=i;  }    }          /*找不到最小的dist，跳出循环*/   if(v==-1)break;  visit[v]=1;  for(i=0;i<Graph->Nv;i++){    //遍历v的所有邻接点，如果经过v到其邻接点的距离更短，更新   if(!visit[i]&&dist[v]+Graph->G[v][i]<dist[i]){  dist[i]=dist[v]+Graph->G[v][i];  }     }  }      /*初始化为下次做准备*/   for(i=0;i<Graph->Nv;i++){  if(dist[i]==INFINITY)  //根据题目要求，无法到达点的距离变为-1dist[i]=-1;  if(Graph->G[S][i]>0)    //初始化dist1，和dist一样dist1[i]=Graph->G[S][i];  if(dist1[i]==dist[i])  /*如果S到i的直接路径为最短路径，更新最短路径条数 */            count[i]=1;  visit[i]=0;  }  visit[S]=1; dist1[S]=0;   count[S]=1;  while(1){  /*第二次dijikstra，计算最短路径条数*/int v=-1;  int min=INFINITY;  for(i=0;i<Graph->Nv;i++){  if(!visit[i]&&dist1[i]<min){  min=dist1[i];  v=i;  }  }  if(v==-1)break;  visit[v]=1;  for(i=0;i<Graph->Nv;i++){  if(!visit[i]&&dist1[v]+Graph->G[v][i]<dist1[i]){  dist1[i]=dist1[v]+Graph->G[v][i];      count[i]=count[v];   //如果经过v到达i的路径为第一条最短路径， 那么从S到i的最短路径条数等于从S到v的最短路径条数                   }  else if(dist1[v]+Graph->G[v][i]==dist1[i])  count[i]+=count[v];      //如果经过v到达i的路径为新的最短路径，那么从S到i的最短路径条数等于从S到i的最短路径条数 加上从S到v的最短路径条数                                                           }      }  }