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。。

 1 #include <bits/stdc++.h>

 2 using namespace std;

 3 const int maxn = 1100;

 4 int n;

 5 double R;

 6 struct Point {

 7     double x,y;

 8 } p[maxn];

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

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

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

12     return sqrt(tmp);

13 }

14 vector<Point>ans;

15 int main() {

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

17         for(int i = 0; i <= n; ++i)

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

19         int low = 0,high = n;

20         Point now = p[low++];

21         ans.clear();

22         double hg = 0;

23         while(low <= high) {

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

25             if((ds - R + hg) > 1e-60) {

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

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

28                 hg += R - hg;

29             }else{

30                 hg += ds;

31                 now = p[low++];

32             }

33             if(fabs(hg - R) < 1e-12) {

34                 ans.push_back(now);

35                 hg = 0.0;

36             }

37         }

38         if(ans.size()) {

39             for(int i = 0; i < ans.size(); ++i)

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

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

42     }

43     return 0;

44 }

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