ZOJ3414 Trail Walk 简单计算几何

ZOJ3414 Trail Walk 简单计算几何

把在n条线段中（以（0,0）为起点）放入m个点，使其等分为m+1份。
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3414

#include < stdio.h >
#include < math.h >
struct  point {
    double x,y,len;
} p[ 1002 ];
int  main()
{
    int n,m,i,j,ca=1;
    double ave,len,ax,ay;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        p[0].x=0; p[0].y=0; ave=0;
        for(i=1;i<=n;i++){
            scanf("%lf%lf",&p[i].x,&p[i].y);
            p[i].len=sqrt((p[i].x-p[i-1].x)*(p[i].x-p[i-1].x)+(p[i].y-p[i-1].y)*(p[i].y-p[i-1].y));
            //printf("%.3f %.3f %.3f\n",p[i].x,p[i].y,p[i].len);
            ave+=p[i].len;
        }
        ave/=(m+1); //每一份的均长
        j=0;  //沿着线段走
       // printf("%.3f\n",ave);
        printf("Route %d\n",ca++);
        for(i=1;i<=m;i++)
        {
            len=0;
            for(; j<n;j++)
            {
                if(len+p[j+1].len>ave)
                    break;
                len+=p[j+1].len;
            }
            double res=ave-len;
            //printf("%d %.3f\n",j,res);
            ax=p[j].x+(res/p[j+1].len)*(p[j+1].x-p[j].x);
            ay=p[j].y+(res/p[j+1].len)*(p[j+1].y-p[j].y);
            printf("CP%d: (%.3lf, %.3lf)\n",i,ax,ay);
            p[j].x=ax; p[j].y=ay;
            p[j+1].len=sqrt((p[j+1].x-p[j].x)*(p[j+1].x-p[j].x)+(p[j+1].y-p[j].y)*(p[j+1].y-p[j].y));
        }
    }
    return 0;
}