hdu 3320 openGL(�I�间变换+矩阵变换)

wangs — Sun, 15 Jul 2012 05:09:00 GMT

http://acm.hdu.edu.cn/showproblem.php?pid=3320

题意�Q?br />      �l�定(x,y,z) 然后3个变换。求最后的�l�果�?br />      初始�?x,y,z)最后给出，变换的顺序由后往前�?br />
我采取的是从后往前做的。还可以使用矩阵�q�算�Q�从前往后，右乘哦�?nbsp;    参考：http://fayaa.com/code/view/9510/

�q�有��是数据的读�?sscanf()是个好东西呀�Q�哈哈�?br />
�ȝ��Q?br />      数学很重要啊�Q?br />

#include<stdio.h>
#include<string.h>
#include<math.h>
struct tra{
    char cm;
    double w;
    double x,y,z;
} p[104];
char ch[125];
int main()
{
    int t;
    int n;
    int i,j,k;
    scanf("%d",&t);
    getchar();
    while (t--)
    {
        gets(ch);
        n=0;
        while (gets(ch)&&ch[2]!='E')
        {
            n++;
            char s[125];
            memset(s,0,sizeof(s));
            if (ch[2]=='R')
            {
                p[n].cm=ch[2];
                i=0;
                while (ch[i]!='(')  i++;
                k=i+1;
                while (ch[i]!=')')  i++;
                j=i-1;
                for (i=k;i<=j;i++)
                    s[i-k]=ch[i];
            //    puts(s);
                sscanf(s,"%lf,%lf,%lf,%lf",&p[n].w,&p[n].x,&p[n].y,&p[n].z);
            }
            else
            {
                i=0;
                while (ch[i]!='(')  i++;
                k=i+1;
                while (ch[i]!=')')  i++;
                j=i-1;
                for (i=k;i<=j;i++)
                    s[i-k]=ch[i];
          //      puts(s);
                p[n].cm=ch[2];
                sscanf(s,"%lf,%lf,%lf",&p[n].x,&p[n].y,&p[n].z);
            }
        }
  //      for (i=n;i>=1;i--)
   //         printf("%c %lf %lf %lf %lf\n",p[i].cm,p[i].w,p[i].x,p[i].y,p[i].z);

        double x,y,z;
//       x=p[n].z,y=p[n].y,z=p[n].z;
        for (i=n;i>=1;i--)
        {
            char c=p[i].cm;
            if (c=='V')
                x=p[i].x,y=p[i].y,z=p[i].z;
            else
                if (c=='S')
                    x *=p[i].x,y *=p[i].y,z *=p[i].z;
                else
                    if (c=='T')
                        x +=p[i].x,y +=p[i].y,z +=p[i].z;
                    else
                    {
                        double si=sin(p[i].w),co=cos(p[i].w);
                        double m=sqrt(p[i].x*p[i].x+p[i].y*p[i].y+p[i].z*p[i].z);
                        double x1,y1,z1;
                        double x2,y2,z2;
                        x1=p[i].x/m;y1=p[i].y/m;z1=p[i].z/m;
                        x2=(x1 * x1 * (1 - co) + co)*x+ (x1 * y1 * (1 - co) - z1 * si)*y+ (x1 * z1 * (1 - co) + y1 * si)*z;
                        y2=(y1 * x1 * (1 - co) + z1 * si)*x+ (y1 * y1 * (1 - co) + co)*y+ (y1 * z1 * (1 - co) - x1 * si)*z;
                        z2=(x1 * z1 * (1 - co) - y1 * si)*x+ (y1 * z1 * (1 - co) + x1 * si)*y+ (z1 * z1 * (1 - co) + co)*z;
                        x=x2;y=y2;z=z2;
                    }
        }
        printf("%.1lf %.1lf %.1lf\n",x,y,z);
    }
    return 0;
}

wangs 2012-07-15 13:09 发表评论

ural 1020-凸包问题

wangs — Mon, 07 May 2012 09:32:00 GMT

acm.timus.ru/problem.aspx?space=1&num=1020

题意是给定一个凸多边形，每个��点有钉子。每个钉子都有半径r。问围绕该凸形的�U�段的长度�?br />
伤心一晚都没有��L��楚题意啊�Q�！�Q�还以�ؓ要先求出凸包呢。。。。。。。�?br />哎，读题是个伤心事�?br />

#include<stdio.h>
#include<string.h>
#include<math.h>
int main()
{
    int n,i;
    double r,x,y,x0,y0,px,py,sum;
    scanf("%d%lf",&n,&r);
    sum=0.0;
    scanf("%lf%lf",&x0,&y0);
    px=x0;py=y0;n--;
    while (n--)
    {
        scanf("%lf%lf",&x,&y);
        sum+=sqrt(pow(x-px,2)+pow(y-py,2));
        px=x;py=y;
    }
    sum+=sqrt(pow(px-x0,2)+pow(py-y0,2));
    sum+=acos(-1.0) * 2 * r;
    printf("%.2lf",sum);
    return 0;
}

wangs 2012-05-07 17:32 发表评论

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hdu 3320 openGL(�I�间变换+矩阵变换)

ural 1020-凸包问题