6-17 Shortest Path [2](25 point(s))(dijkstra)

本文介绍了一个使用Dijkstra算法来寻找有向图中从指定源节点到所有其他节点的加权最短距离的程序实现。通过初始化距离数组并采用贪心策略,该算法逐步确定每个顶点的最短路径。

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6-17 Shortest Path [2](25 point(s))

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

题意有点小问题,实际上是求s到其他各点的距离,迪杰斯特拉就可以
code:
//迪杰斯特拉算法求最短路
void ShortestDist( MGraph Graph, int dist[], Vertex S ){
    int book[MaxVertexNum];//标记数组
    int i,j,k;
    for(j = 0; j < Graph->Nv; j++){
        book[j] = 0;
        dist[j] = Graph->G[S][j];
    }//初始化
    book[S] = 1;
    dist[S] = 0;//一开始忘了初始化起点,导致一直错
    for(j = 0; j < Graph->Nv-1; j++){//迪杰斯特拉算法
        int min = INFINITY;
        int u = -1;
        for(k = 0; k < Graph->Nv; k++){
            if(book[k]==0&&dist[k]<min){//找到最小值
                min = dist[k];
                u = k;
            }
        }
        if(u==-1)break;//如果u不变说明没有可以松弛的边了
        book[u] = 1;//加入,表示完成松弛的边
        for(k = 0; k < Graph->Nv; k++){//以刚找到的值为中间点判断能否继续松弛
            if(!book[k]&&dist[k]>dist[u]+Graph->G[u][k]){//注意!!这里判断条件
                dist[k]=dist[u]+Graph->G[u][k];//必须必须有book[k]==0这个,否则是段错误
            }
            
        }
    }
    for(i = 0; i < Graph->Nv; i++){
        if(dist[i]==INFINITY)dist[i] = -1;//无穷大赋值为-1
    }
}

 
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