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

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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 90 1 10 5 10 6 15 3 12 1 22 6 36 4 44 5 56 5 122

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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