[LightOJ1240]Point Segment Distance 题解

题意简述

原题LightOJ 1240，Point Segment Distance(3D)。

求三维空间里线段AB与C。

题解

我们设一个点在线段AB上移动，然后发现这个点与原来的C点的距离呈一个单峰状，于是珂以三分。

三分的时候处理一下，把A点坐标移动到(0,0,0)珂以方便操作。

代码

#include <cstdio>
#include <cmath> 

namespace fast_IO{
    const int IN_LEN = 10000000, OUT_LEN = 10000000;
    char ibuf[IN_LEN], obuf[OUT_LEN], *ih = ibuf + IN_LEN, *oh = obuf, *lastin = ibuf + IN_LEN, *lastout = obuf + OUT_LEN - 1;
    inline char getchar_(){return (ih == lastin) && (lastin = (ih = ibuf) + fread(ibuf, 1, IN_LEN, stdin), ih == lastin) ? EOF : *ih++;}
    inline void putchar_(const char x){if(oh == lastout) fwrite(obuf, 1, oh - obuf, stdout), oh = obuf; *oh ++= x;}
    inline void flush(){fwrite(obuf, 1, oh - obuf, stdout);}
    int read(){
        int x = 0; int zf = 1; char ch = ' ';
        while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar_();
        if (ch == '-') zf = -1, ch = getchar_();
        while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar_(); return x * zf;
    }
    void write(int x){
        if (x < 0) putchar_('-'), x = -x;
        if (x > 9) write(x / 10);
        putchar_(x % 10 + '0');
    }
}

using namespace fast_IO;

struct Pos{
	double x, y, z;
} pos[3];

long double cacDis(Pos &a, Pos &b){
	return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y) + (a.z - b.z) * (a.z - b.z));
}

long double cac(long double num){
	Pos _new = (Pos){pos[1].x * num, pos[1].y * num, pos[1].z * num};
	Pos mid = (Pos){(_new.x + pos[0].x) / 2, (_new.y + pos[0].y) / 2, (_new.z + pos[0].z) / 2};
	return cacDis(mid, pos[2]);
}

int main(){
	int t = read();
	for (int j = 1; j <= t; ++j){
		pos[0].x = read(), pos[0].y = read(), pos[0].z = read();
		pos[1].x = read(), pos[1].y = read(), pos[1].z = read(	);
		pos[1].x -= pos[0].x, pos[1].y -= pos[0].y, pos[1].z -= pos[0].z;
		pos[2].x = read(), pos[2].y = read(), pos[2].z = read();
		pos[2].x -= pos[0].x, pos[2].y -= pos[0].y, pos[2].z -= pos[0].z;
		pos[0].x -= pos[0].x, pos[0].y -= pos[0].y, pos[0].z -= pos[0].z;
		long double l = 0, r = 2.0, mid;
		long double mmid; long double ans = 0.0;
		for (int i = 1; i <= 100; ++i){
			mid = (l + r) / 2;
			mmid = (mid + r) / 2;
			if (cac(mid) <= cac(mmid))
				r = mmid, ans = mmid;
			else
				l = mid;
		}
		printf("Case %d: %.9Lf\n", j, cac(ans));
	}
	return 0;
}