1549: Navigition Problem （几何计算+模拟细节较多）

1549: Navigition Problem

Submit Page Summary Time Limit: 1 Sec Memory Limit: 256 Mb Submitted: 400 Solved: 122

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

题目意思：
给你很多点，将这些点连成线段，起点和终点不相连
从起点出发，沿着这些折线走，每一次走d距离
问你每走d距离达到的所有点的坐标
（包括起点和终点）

分析：
就是几何计算+模拟，细节多，每沿着折线走d距离，就是输出一下到达该点的坐标
总长度不足d
就按照题目输出字符串
下一步走d距离超过了终点的话
最后只要输出终点
不用输出超出终点的点

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
using namespace std;
typedef long long LL;
#define max_v 1005
struct node
{
    double x,y;
}p[max_v];
double len[max_v];
double dis(node a,node b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double sum[max_v];
int main()
{
    int n;
    double d;
    while(~scanf("%d %lf",&n,&d))
    {
        double c=0;
        sum[0]=0;
        for(int i=1;i<=n+1;i++)
        {
            scanf("%lf %lf",&p[i].x,&p[i].y);
            if(i>1)
            {
                len[i-1]=dis(p[i],p[i-1]);
                c+=len[i-1];
                sum[i-1]=sum[i-2]+len[i-1];
            }
        }
        if(c<d)
        {
             printf("No Such Points.\n");
             continue;
        }
        double now=0,k,x,y;
        for(int i=1;i<=n;)
        {
            if(now+d<=sum[i])
            {
                double L=now+d-sum[i-1];
                double x1=p[i].x;
                double y1=p[i].y;
                double x2=p[i+1].x;
                double y2=p[i+1].y;

                if(x1!=x2)
                {
                    k=(y1-y2)/(x1-x2);
                    if(x2<x1)
                    {
                        x=x1-sqrt((L*L)/(k*k+1));
                        y=k*(x-x1)+y1;
                    }else
                    {
                        x=x1+sqrt((L*L)/(k*k+1));
                        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++;
            }
        }
    }
    return 0;
}
/*
题目意思：
给你很多点，将这些点连成线段，起点和终点不相连
从起点出发，沿着这些折线走，每一次走d距离
问你每走d距离达到的所有点的坐标
（包括起点和终点）

分析：
就是几何计算+模拟，细节多，每沿着折线走d距离，就是输出一下到达该点的坐标
总长度不足d
就按照题目输出字符串
下一步走d距离超过了终点的话
最后只要输出终点
不用输出超出终点的点

*/

原文地址：https://www.cnblogs.com/yinbiao/p/9498593.html