ZOJ 1203 Swordfish

prim实现 注意segmentation fault

#include <cstdio>
#include <cmath>

double dis[105][105];
double p[105][2]; //输入坐标点
double d[105];  //一点到另一点的距离
double vis[105];  //记得该点是否已被访问
double len;
int n;

//初始化dis[105][105]
void init()
{
    int i,j;
    for(i=1; i<=n; i++)
    {
        for(j=1; j<=n; j++)
        {
            dis[i][j]=dis[j][i]=100000000;
        }
        vis[i] = 0;
    }
    len=0.0;
}

//计算dis[105][105]
void makedis()
{
    int i,j;

    for(i = 1; i <= n; i++)
        for(j = i; j <= n; j++)
        {
            dis[i][j] = dis[j][i] = sqrt((p[i][0]-p[j][0])*(p[i][0]-p[j][0])+(p[i][1]-p[j][1])*(p[i][1]-p[j][1]));

        }
}


//用prim算法求最短距离
void prim()
{
    int i,j,k;
    double min;
    for(i = 1; i <= n; i++)
        d[i] = dis[1][i];
    vis[1] = 1;

    for( i = 1; i < n; i++)
    {
        min = 0xfffffff;              // 最大值
        for ( j = 1; j <= n; j++)
        {
            if ( d[j] < min && !vis[j])
            {
                min = d[j];
                k = j;
            }
        }
        vis[k] = 1;
        len += d[k];
        for(j = 1; j <= n; j++)
        {
            if(!vis[j] && d[j] > dis[k][j])
                d[j] = dis[k][j];
        }

    }
}

//主函数
int main()
{
    int step;
    int i;
    step=0;
    while(scanf("%d",&n) && n)     //以0结束
    {
        step++;              //记录case

        if(step!=1)
            printf("
");
        init();
        for(i = 1; i <= n; i++)
        {
            scanf("%lf%lf",&p[i][0],&p[i][1]);
        }
        makedis();
        prim();
        printf("Case #%d:
",step);
        printf("The minimal distance is: %.2lf
",len);
    }
    return 0;
}