原题LightOJ 1240，Point Segment Distance(3D)。
求三维空间里线段AB与C。
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。
#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; }