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

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
    }
}