K - Keeping the Dogs Apart GYM-101550（計算幾何）

參考：https://blog.csdn.net/xbb224007/article/details/79846472

題意：
給出兩隻狗的路線，且其速度相同，求其運動時候的最短距離。

思路：
對於兩個相同長度的向量，其距離按照時間爲自變量構成一個二次函數。
那麼對於兩隻狗每次取相同長度的直線來計算距離，然後推下去。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <cmath>
using namespace std;

const int maxn = 1e5 + 7;

struct Point {
    double x,y;
}p1[maxn],p2[maxn];

double dis(Point a,Point b) {
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

Point get(Point a,Point b,double l) { //獲取向量(a,b)長度爲l的一段
    Point ans;
    double len = dis(a,b);
    double ex = (b.x - a.x) / len,ey = (b.y - a.y) / len;
    ans.x = a.x + ex * l;
    ans.y = a.y + ey * l;
    return ans;
}

double solve(Point a1,Point b1,Point a2,Point b2) { //計算最短距離
    double L1 = dis(a1,b1),L2 = dis(a2,b2);
    double ex1 = (b1.x - a1.x) / L1,ey1 = (b1.y - a1.y) / L1;
    double ex2 = (b2.x - a2.x) / L2,ey2 = (b2.y - a2.y) / L2;
    
    double a = (ex1 - ex2) * (ex1 - ex2) + (ey1 - ey2) * (ey1 - ey2);
    double b = 2 * ((a1.x - a2.x) * (ex1 - ex2) + (a1.y - a2.y) * (ey1 - ey2));
    double c = (a1.x - a2.x) * (a1.x - a2.x) + (a1.y - a2.y) * (a1.y - a2.y);
    
    if(a == 0) {
        return min(dis(a1,a2),dis(b1,b2));
    }
    double x = -b / (2 * a);
    if(x < 0 || x > L1) {
        return min(dis(a1,a2),dis(b1,b2));
    }
    return sqrt(a * x * x + b * x + c);
}

int main() {
    int n;scanf("%d",&n);
    for(int i = 1;i <= n;i++) {
        scanf("%lf%lf",&p1[i].x,&p1[i].y);
    }
    int m;scanf("%d",&m);
    for(int i = 1;i <= m;i++) {
        scanf("%lf%lf",&p2[i].x,&p2[i].y);
    }
    
    int i = 2,j = 2;
    double mi = 1e9;
    while(i <= n && j <= m) {
        double l1 = dis(p1[i - 1],p1[i]),l2 = dis(p2[j - 1],p2[j]);
        if(fabs(l1 - l2) < 1e-8) {
            mi = min(mi,solve(p1[i - 1],p1[i],p2[j - 1],p2[j]));
            i++;j++;
        }
        else if(l1 < l2) {
            Point now = get(p2[j - 1],p2[j],l1);
            mi = min(mi,solve(p1[i - 1], p1[i], p2[j - 1], now));
            i++;
            p2[j - 1] = now;
        }
        else if(l1 > l2) {
            Point now = get(p1[i - 1],p1[i],l2);
            mi = min(mi,solve(p1[i - 1],now,p2[j - 1],p2[j]));
            j++;
            p1[i - 1] = now;
        }
    }
    
    printf("%.10f\n",mi);
    return 0;
}