6-17 Shortest Path [2](25 分)

Write a program to find the weighted shortest distances 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[], 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.

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[], Vertex S );

int main()
{
    int dist[MaxVertexNum];
    Vertex S, V;
    MGraph G = ReadG();

    scanf("%d", &S);
    ShortestDist( G, dist, S );

    for ( V=0; V<G->Nv; V++ )
        printf("%d ", dist[V]);

    return 0;
}

/* Your function will be put here */

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

7 9
0 1 1
0 5 1
0 6 1
5 3 1
2 1 2
2 6 3
6 4 4
4 5 5
6 5 12
2

Sample Output:

-1 2 0 13 7 12 3


不能抵达的为INFINITY,用过dijkstra算法,最后记得把INFINITY变成-1,dist[S]变成0
代码:
void ShortestDist( MGraph Graph, int dist[], Vertex S )
{
    for(int i = 0;i < Graph -> Nv;i ++)
    dist[i] = Graph -> G[S][i];
    int vis[MaxVertexNum] = {0};
    vis[S] = 1;
    for(int i = 0;i < Graph -> Nv;i ++)
    {
        int min = INFINITY;
        int t = INFINITY;
        for(int j = 0;j < Graph -> Nv;j ++)
            if(!vis[j]&&dist[j] < min)min = dist[j],t = j;
        if(min == INFINITY)continue;
        vis[t] = 1;
        for(int j = 0;j < Graph -> Nv;j ++)
        {
            if(!vis[j])
            {
                if(dist[j] > Graph -> G[t][j] + min)dist[j] = Graph -> G[t][j] + min;
            }
        }
    }
    for(int i = 0;i < Graph -> Nv;i ++)
    if(i == S)dist[i] = 0;
    else if(dist[i] == INFINITY)dist [i] = -1;
}
原文地址:https://www.cnblogs.com/8023spz/p/7756638.html