关于退火有关问题的一段C代码，求修改

关于退火问题的一段C代码，求修改！
其中有些东西是我调试的时候用的！求大神帮忙修改修改！

C/C++ code

#include <stdio.h>#include <stdlib.h>#include <math.h>#include <time.h>#define Max_distance 9999999#define N 30double x[N]={0.37,0.75,0.45,0.76,0.71,0.07,0.42,0.59,0.32,0.60,0.30,0.67,0.62,0.67,0.20,0.35,0.27,0.94,0.82,0.37,0.61,0.42,0.60,0.39,0.53,0.40,0.63,0.50,0.98,0.68};double y[N]={0.91,0.87,0.85,0.75,0.72,0.74,0.71,0.69,0.64,0.64,0.59,0.59,0.55,0.55,0.50,0.45,0.43,0.42,0.38,0.27,0.26,0.25,0.23,0.19,0.19,0.13,0.08,0.04,0.02,0.85};double dis[N][N]={0};int path[N];void test(){    int i,j;    printf("\n\n");    for(i=0;i<N;i++)        printf("%d ",path[i]);/*    for(i=0;i<N;i++)    {        for(j=0;j<N;j++)        {            printf("%d ",dis[i][j]);            if(j+1%6==0)                printf("\n");        }        printf("\n");printf("\n");printf("\n");printf("\n");    }*/    printf("\n\n");}//计算距离void count_distance(){    int i,j;    for(i=0;i<N;i++)    {        for(j=0;j<N;j++)        {            dis[i][j]=sqrt(pow((x[i]-x[j]),2)+pow((y[i]-y[j]),2));            printf("%f ",dis[i][j]);            if(j+1%6==0)                printf("\n");        }        printf("\n");printf("\n");printf("\n");    }}//初始化一条路void init_path(){    int i;    for(i=0;i<N;i++)    {        path[i]=i;        printf("%d ",path[i]);    }}//计算路程double count_lucheng(int a,int b){    double s=0;    int i;    if(a<0)        a=0;    if(b>=N-1)        b=N-1;    for(i=a;i<b;i++)    {        s+=dis[path[i]][path[i+1]];    }    return s;}//退火算法double tuihuo(){    init_path();    //退火条件    test();    double T=1;    double at=0.999,df;    double e=pow(0.1,30);    int L=20000;    //开始退火    srand(time(0));        //计算初始路程    double sum=count_lucheng(0,N-1);        int i=0;int a,b;    while(i<L)    {        a=rand()%30;        b=rand()%30;        df=dis[a-1][b]+dis[a][b+1]-dis[a-1][a]-dis[b][b+1];        double d=exp(-df/T);        int c=rand();        if(df<0||(df>=0&&(d>c)))//接受新路径        {            int newpath[N];            int j;                        int mm=(a>b? a:b);            int nn=(a<b? a:b);            int k=nn;            for(j=nn;j<=mm;j++)            {                newpath[j]=path[j];            }            for(j=mm;j>=nn;j--)            {                path[k]=newpath[j];                k=k+1;            }            sum+=df;        }        T=T*at;        if(T<e)            break;        i++;        test();        printf("%f ",sum);        float newdis=count_lucheng(0,N-1);        printf("%f ",newdis);    }    return sum;}void main(){    count_distance();    double www=tuihuo();    //test();    int i;    for(i=0;i<N;i++)    {        printf("%d    ",path[i]);    }    printf("%f\n",www);}

C/C++ code

 while(i<L)    {        a=rand()%30;        b=rand()%30;        df=dis[a-1][b]+dis[a][b+1]-dis[a-1][a]-dis[b][b+1];        double d=exp(-df/T);        int c=rand();        if(df<0||(df>=0&&(d>c)))//接受新路径//这里df《0  sum后面就会变小咯        {            int newpath[N];            int j;                        int mm=(a>b? a:b);            int nn=(a<b? a:b);            int k=nn;            for(j=nn;j<=mm;j++)            {                newpath[j]=path[j];            }            for(j=mm;j>=nn;j--)            {                path[k]=newpath[j];                k=k+1;            }            sum+=df;        }
 
 [解决办法]
        a=rand()%30;
       b=rand()%30;
       df=dis[a-1][b]+dis[a][b+1]-dis[a-1][a]-dis[b][b+1];

当a为0时，dis[a-1][b]和dis[a-1][a]越界。当b为29时，dis[a][b+1]和dis[b][b+1]越界。