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 } }