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 = 1100;
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 = 0; i <= n; ++i)
            scanf("%lf %lf",&p[i].x,&p[i].y);
        int low = 0,high = n;
        Point now = p[low++];
        ans.clear();
        double hg = 0;
        while(low <= high) {
            double ds = getDis(now,p[low]);
            if((ds - R + hg) > 1e-60) {
                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-12) {
                ans.push_back(now);
                hg = 0.0;
            }
        }
        if(ans.size()) {
            for(int i = 0; i < ans.size(); ++i)
                printf("%.2f,%.2f
",ans[i].x,ans[i].y);
        } else puts("No Such Points.");
    }
    return 0;
}