无向图最短路径和递归调用

Question

这是一个球无向图最短路径的代码，所有路径圈值都为1
点二行代码
#define POINT_COUNT_MAX 7
改成#define POINT_COUNT_MAX 8
运行就会崩溃，不知道为什么，个人猜测可能是递归调用的太多
求解答

 #include<stdio.h>
#define POINT_COUNT_MAX     7
#define END_INDEX           17

int flag = 1;
int allPath[100][POINT_COUNT_MAX];
int pathNum = 0;

typedef struct POINT
{
    int pathCount;
    int path[20];
}POINT;

void fun(POINT *point, int index, int *pointCount);

void fun(POINT *point, int index, int *pointCount)
{
    int i = 0;
    int temp;
    allPath[pathNum][(*pointCount)-1] = index;

        if(*pointCount >= POINT_COUNT_MAX || index == END_INDEX)
        {
            if(index == END_INDEX)
            {
                pathNum++;

                for(temp = 0; temp < POINT_COUNT_MAX; temp++)
                    allPath[pathNum][temp] = allPath[pathNum-1][temp];
            }
        }else{
            for(i = 0; i < point[index].pathCount; i++)
            {
                *pointCount = *pointCount + 1;
                fun(point, *(point[index].path + i), pointCount); 
                *pointCount = *pointCount - 1;
            }
        } 
}

void main(void)
{
    int pointCount = 1;
    int i, j;

    POINT point[18] = 
    {
    /*0*/   {3,1,2,3},
    /*1*/   {3,2,4,9,},
    /*2*/   {4,1,3,4,5},
    /*3*/   {4,2,5,6,7},
    /*4*/   {5,1,2,5,8,9},
    /*5*/   {7,2,3,4,6,9,10,12},
    /*6*/   {7,3,5,7,8,12,13,14},
    /*7*/   {3,3,6,8},
    /*8*/   {4,6,7,14,15},
    /*9*/   {5,1,4,5,10,11},
    /*10*/  {4,5,9,11,12},
    /*11*/  {3,9,10,16},
    /*12*/  {5,5,6,10,13,16},
    /*13*/  {6,6,12,14,15,16,END_INDEX},
    /*14*/  {4,6,8,13,15},
    /*15*/  {4,8,13,14,END_INDEX},
    /*16*/  {4,11,12,13,END_INDEX},
    /*17*/  {0,0}
    };
    fun(point, 0, &pointCount); 

    for(i = 0; i < pathNum; i++)
    {
        for(j = 0; j < POINT_COUNT_MAX; j++)
            printf("%d", allPath[i][j]);
        printf("\n");
    }   
}

Answer 1

递归调用的太多，你递归能有多少次。。递归确实很能消耗，但这个程序应该不至于