ZOJ-3414-Trail Walk_计算几何

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我打代码这么狗还这么懒，以后怎么办。

题意：

给n个点的坐标（整数），以坐标系原点为起点，依次连接各个点，给出正整数m，要求添加m个CP点（check point），把折线按长度平均分开（把起点考虑为第0个和第m+1个CP点）。

Input

There are multiple test cases. The first line of each case contains two integer n, m(1 <= n, m <= 1000). Then n pairs of integer followed, giving the coordinate of p_i.

Output

The first line of each case, output "Route case_number", Then m lines followed, the ith line contains "CPcase_num: (x_i, y_i)" where (x_i, y_i) represent the coordinate of the CP_i. Always keep three number after the decimal point.

Sample Input

3 31 12 33 5

Sample Output

Route 1CP1: (1.026, 1.051)CP2: (1.684, 2.368)CP3: (2.342, 3.684)

计算几何里应该不难的题吧，但是由于我计算几何就是狗，而且有点新感悟就写一下。

大体过程只要按正常想法就能模拟出来，关键是公式，计算几何不要直接使用斜率的数值，但是可以通过把k写成(y1-y2)/(x1-x2)的形式表达出来，然后推出最后的式子以后再代数，这样就不会有使用斜率数值的误差了。

代码如下

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>using namespace std;#define mxn 1010int n,m;double len;bool dcmp(double a,double b){return fabs(a-b)<0.001;}struct point{double x,y;point(double _x,double _y){x=_x,y=_y;}point(){};}p[mxn];struct ret{int p_id;double usd;ret(int a,double b){p_id=a,usd=b;}};double dist(point a,point b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}double deltaX(int in,double l){return l*(p[in+1].x-p[in].x)/sqrt((p[in+1].y-p[in].y)*(p[in+1].y-p[in].y)+(p[in+1].x-p[in].x)*(p[in+1].x-p[in].x));}double deltaY(int in,double l){return l*(p[in+1].y-p[in].y)/sqrt((p[in+1].y-p[in].y)*(p[in+1].y-p[in].y)+(p[in+1].x-p[in].x)*(p[in+1].x-p[in].x));}ret find(ret in,double tgt,int cp_id){double tem=dist(p[in.p_id],p[in.p_id+1]);if(tem-in.usd<=tgt||dcmp(tem-in.usd,tgt))return find(ret(in.p_id+1,0.0),tgt-tem+in.usd,cp_id);printf("CP%d: ",cp_id);double delta_x=deltaX(in.p_id,tgt+in.usd);double delta_y=deltaY(in.p_id,tgt+in.usd);printf("(%.3lf, %.3lf)\n",p[in.p_id].x+delta_x,p[in.p_id].y+delta_y);return ret(in.p_id,in.usd+tgt);}int main(){int cs=1;while(scanf("%d%d",&n,&m)!=EOF){double sum=0;p[0]=point(0.0,0.0);for(int i=1;i<=n;++i){scanf("%lf%lf",&p[i].x,&p[i].y);sum+=dist(p[i-1],p[i]);}len=sum/(m+1);ret start(0,0.0);printf("Route %d\n",cs++);for(int i=0;i<m;++i){ret tem=find(start,len,i+1);start=tem;}}return 0;}

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