csu 1549: Navigition Problem(几何,模拟)

1549: Navigition Problem

Time Limit: 1 Sec  Memory Limit: 256 MB
Submit: 305  Solved: 90
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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

题意:有 n 条首尾相连的线段 (第1条不和第 n条相连).从第一个点的起点开始每次沿着线段平移 d 个单位到达某一个点，把每次的坐标输出,如果所有线段之和都小于d，输出 "No Such Points."
题解:模拟这个过程,每次确定当前的点在哪条线段上,假设当前点在line[i] ,那么我们设当前点为 (x,y)， 可以得到它与line[i]起点的距离,然后通过公式可以推出当前点的位置，记得讨论斜率不存在的
情况。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <math.h>
#include <algorithm>
using namespace std;
const int N = ;
struct Point
{
    double x,y;
} p[N];
struct Line
{
    Point a,b;
} line[N];
double len[N];
double dis(Point a,Point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double sum[N];
int main()
{
    int n;
    double d;
    while(scanf("%d%lf",&n,&d)!=EOF)
    {
        double SUM = ;
        sum[] = ;
        for(int i=; i<=n+; i++)
        {
            scanf("%lf%lf",&p[i].x,&p[i].y);
            if(i>)
            {
                len[i-] = dis(p[i],p[i-]);
                SUM+=len[i-];
                sum[i-] = sum[i-]+len[i-];
            }
        }
        if(SUM<d)
        {
            printf("No Such Points.\n");
            continue;
        }
        double now = ,k,x,y;
        for(int i=; i<=n;)
        {
            if(now+d<=sum[i])
            {
                double L = now+d - sum[i-];
                double x1 = p[i].x,y1 = p[i].y;
                double x2 = p[i+].x,y2 = p[i+].y;
                if(x1!=x2)
                {
                    k = (y1-y2)/(x1-x2);
                    if(x2<x1)
                    {
                        x = x1-sqrt((L*L)/(k*k+));
                        y = k*(x-x1)+y1;
                    }
                    else
                    {
                        x = x1+sqrt((L*L)/(k*k+));
                        y = k*(x-x1)+y1;
                    }
                }
                else
                {
                    x = x1;
                    if(y2<y1) y = y1-L;
                    else  y = y1+L;
                }
                printf("%.2lf,%.2lf\n",x,y);
                now = now+d;
            }
            else
            {
                i++;
            }
        }

    }
}