标签：count 25 dist int Graph MGraph MaxVertexNum Path Shortest

Write a program to not only find the weighted shortest distances, but also count the number of different minimum paths 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[], int count[], 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. The number of different minimum paths from V to the source S is supposed to be stored in count[V] and count[S]=1.

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[], int count[], Vertex S );

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

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

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

    return 0;
}

/* Your function will be put here */

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

8 11
0 4 5
0 7 10
1 7 30
3 0 40
3 1 20
3 2 100
3 7 70
4 7 5
6 2 1
7 5 3
7 2 50
3

Sample Output:

40 20 100 0 45 53 -1 50 
1 1 4 1 1 3 0 3 

这道题就是Dijkstra算法，是一种贪心算法，分解开来看就是从起始点开始，找离这个顶点集最近的点，然后再更新路径。

唯一有点困难的可能就是数相同的最短路径了。仔细想想，在Dijkstra过程中，这个应当是递增的，如果a到达b的路径有k条，而b能达到c，则a到c的路径也是k条，如果b到c有两条路径，则a到c的路径是k+1条。

根据这个原则，构建代码。

int known[MaxVertexNum];

void ShortestDist(MGraph Graph, int dist[], int count[], Vertex S) {
    for (int i = 0; i < Graph->Nv; i++) dist[i] = INFINITY, count[i] = 0;
    dist[S] = 0, count[S] = 1;

    while (S != -1) {
        known[S] = 1;
        for (int i = 0; i < Graph->Nv; i++) {
            if (!known[i]) {
                if (dist[i] > Graph->G[S][i] + dist[S])
                    dist[i] = Graph->G[S][i] + dist[S], count[i] = count[S];
                else if (dist[i] == Graph->G[S][i] + dist[S])
                    count[i] += count[S];
            }
        }

        S = -1;
        for (int i = 0, v_min = INFINITY; i < Graph->Nv; i++)
            if (!known[i] && v_min > dist[i])
                v_min = dist[i], S = i;
    }

    for (int i = 0; i < Graph->Nv; i++) {
        if (dist[i] == INFINITY) count[i] = 0, dist[i] = -1;
    }
}

标签：count,25,dist,int,Graph,MGraph,MaxVertexNum,Path,Shortest
来源： https://www.cnblogs.com/nonlinearthink/p/12178281.html

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