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

2 0.50
0.00 0.00
1.00 0.00
1.00 1.00

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 ;
 }