CSUOJ 1549 Navigition Problem

1549: Navigition Problem

Time Limit: 1 Sec Memory Limit: 256 MB
Submit: 65 Solved: 12

Description

Navigation is a field of study that focuses on the process of monitoring and controlling the movement of a craft or vehicle from one place to another. The field of navigation includes four general categories: land navigation, marine navigation, aeronautic navigation, and space navigation. (From Wikipedia)
Recently, slowalker encountered a problem in the project development of Navigition APP. In order to provide users with accurate navigation service , the Navigation APP should re-initiate geographic location after the user walked DIST meters. Well, here comes the problem. Given the Road Information which could be abstracted as N segments in two-dimensional space and the distance DIST, your task is to calculate all the postion where the Navigition APP should re-initiate geographic location.

Input

The input file contains multiple test cases.
For each test case,the first line contains an interger N and a real number DIST. The following N+1 lines describe the road information.
You should proceed to the end of file.

Hint:
1 <= N <= 1000
0 < DIST < 10,000

Output

For each the case, your program will output all the positions where the Navigition APP should re-initiate geographic location. Output “No Such Points.” if there is no such postion.

Sample Input

Sample Output

0.50,0.00

1.00,0.00

1.00,0.50

1.00,1.00

HINT

Source

解题：此题太JB难读了，就是每次走了dist距离然后输出此时的坐标位置，记得一定要走够dist的距离才输出坐标，还有坐标中间有逗号。。

 #include <bits/stdc++.h>

 using namespace std;

 const int maxn = ;

 int n;

 double R;

 struct Point {

     double x,y;

 } p[maxn];

 double getDis(const Point &a,const Point &b) {

     double tmp = (a.x - b.x)*(a.x - b.x);

     tmp += (a.y - b.y)*(a.y - b.y);

     return sqrt(tmp);

 }

 vector<Point>ans;

 int main() {

     while(~scanf("%d %lf",&n,&R)) {

         for(int i = ; i <= n; ++i)

             scanf("%lf %lf",&p[i].x,&p[i].y);

         int low = ,high = n;

         Point now = p[low++];

         ans.clear();

         double hg = ;

         while(low <= high) {

             double ds = getDis(now,p[low]);

             if((ds - R + hg) > 1e-) {

                 now.x += (p[low].x - now.x)/ds*(R-hg);

                 now.y += (p[low].y - now.y)/ds*(R-hg);

                 hg += R - hg;

             }else{

                 hg += ds;

                 now = p[low++];

             }

             if(fabs(hg - R) < 1e-) {

                 ans.push_back(now);

                 hg = 0.0;

             }

         }

         if(ans.size()) {

             for(int i = ; i < ans.size(); ++i)

                 printf("%.2f,%.2f\n",ans[i].x,ans[i].y);

         } else puts("No Such Points.");

     }

     return ;

 }