for(i=0;i<n;i++)  
    {
     for(j=0;j<n;j++)
        {if (path[i][j]!=i)  
           {
              i1=path[i][j];
              printf("%d %d %d\n",i,i1,j);
              x[i1][j]=x[i1][j]+b[i][j];
           }
              printf("%d\n",i1);
          loop:if (path[i][i1]!=i)                 
           {i11=path[i][i1];
            x[i11][i1]=x[i11][i1]+b[i][j];
            i1=path[i][i11];goto loop;
           }
          // printf("%d",path[i][i1]);
          if(path[i][j]==i)
           {
            x[i][j]=x[i][j]+b[i][j];
           }
             printf("%d %d/n",i,j);
     }
     }
这个只能执行第一个if，然后系统会报错，请帮忙看看该怎么改，谢谢。

[ 本帖最后由 shenlan4939 于 2014-3-12 10:58 编辑 ]

这个是要干什么，问问题要全面啊

这是全部的程序，是全有全无配流的代码。但程序只能执行第一个if语句，以后就会报错
#include<stdio.h>
#define N 9
#define a 10000
void main()
{
    int n=9;
    int path[9][9]={-1,-1,1,-1,3,4,3,4,5,-1,-1,-1,-1,-1,4,-1,4,5,-1,-1,-1,-1,-1,-1,-1,-1,5,-1,-1,-1,-1,-1,4,-1,4,5,-1,-1,-1,-1,-1,-1,-1,-1,5,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,7,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1};
    int x[9][9]={0},b[9][9]={0,0,0,0,0,0,0,0,1000,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};
  
int i,j,i1,i11;
     for(i=0;i<n;i++)
    {
     for(j=0;j<n;j++)
        {if (path[i][j]!=i)
           {
              i1=path[i][j];
              printf("%d %d %d\n",i,i1,j);
              x[i1][j]=x[i1][j]+b[i][j];
           }
              printf("%d\n",i1);
          loop:if (path[i][i1]!=i)               
           {i11=path[i][i1];
            x[i11][i1]=x[i11][i1]+b[i][j];
            i1=path[i][i11];goto loop;
           }
          // printf("%d",path[i][i1]);
          if(path[i][j]==i)
           {
            x[i][j]=x[i][j]+b[i][j];
           }
             printf("%d %d/n",i,j);
     }
     }
printf("路段流量矩阵为:\n");
for(i=0;i<n;i++)
    {for(j=0;j<n;j++)
    printf("%4d",x[i][j]);
    printf("\n");}  

}


[ 本帖最后由 shenlan4939 于 2014-3-12 10:59 编辑 ]

loop:if (path[i][i1]!=i)   // 数组下标什么时候有-1了??