标签:acm
ACM
题目地址:
HDU 3272 - Mission Impossible
题意:
在二维平面上,给你一个初始位置(hx,hy),你需要获得四种资源,A在x轴上任意位置,B在y轴上任意位置,C、D位置会告诉你。问获得四种资源后返回(hx,hy)最短要走多长。
分析:
三条线段与X、Y轴相交有三种情况:
代码:
/* * Author: illuz <iilluzen[at]gmail.com> * Blog: http://blog.csdn.net/hcbbt * File: 3272.cpp * Create Date: 2014-08-15 20:56:16 * Descripton: */ #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; #define repf(i,a,b) for(int i=(a);i<=(b);i++) typedef long long ll; const int N = 0; struct Point { double x; double y; } h, o, p[3]; double dis(const Point& a, const Point& b) { return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y)); } // calc h->a->b->h double calc(int a, int b) { double ret = dis(h, p[a]) + dis(p[a], p[b]) + dis(p[b], h); double ha, ab, bh; bool cx = 0, cy = 0; if (h.x * p[a].x <= 0 || p[a].x * p[b].x <= 0 || p[b].x * h.x <= 0) cx = 1; if (h.y * p[a].y <= 0 || p[a].y * p[b].y <= 0 || p[b].y * h.y <= 0) cy = 1; if (cx == 0 && cy == 0) { // 3 Point t1, t2; t1.x = h.x; t1.y = -h.y; t2.x = -p[a].x; t2.y = p[a].y; ha = ret - dis(h, p[a]) + dis(t1, t2); t1.x = p[b].x; t1.y = -p[b].y; t2.x = -p[a].x; t2.y = p[a].y; ab = ret - dis(p[a], p[b]) + dis(t1, t2); t1.x = p[b].x; t1.y = -p[b].y; t2.x = -h.x; t2.y = h.y; bh = ret - dis(h, p[b]) + dis(t1, t2); double ans = min(ha, min(ab, bh)); // 6 // ha->x ab->y t1.x = p[a].x; t1.y = -p[a].y; t2.x = -p[a].x; t2.y = p[a].y; ans = min(ans, ret - dis(p[a], h) + dis(t1, h) - dis(p[a], p[b]) + dis(t2, p[b])); // ha->x bh->y t1.x = h.x; t1.y = -h.y; t2.x = -h.x; t2.y = h.y; ans = min(ans, ret - dis(p[a], h) + dis(p[a], t1) - dis(h, p[b]) + dis(t2, p[b])); // ab->x ha->y t1.x = p[a].x; t1.y = -p[a].y; t2.x = -p[a].x; t2.y = p[a].y; ans = min(ans, ret - dis(p[a], p[b]) + dis(t1, p[b]) - dis(p[a], h) + dis(t2, h)); // ab->x bh->y t1.x = p[b].x; t1.y = -p[b].y; t2.x = -p[b].x; t2.y = p[b].y; ans = min(ans, ret - dis(p[a], p[b]) + dis(t1, p[a]) - dis(p[b], h) + dis(t2, h)); // bh->x ha->y t1.x = h.x; t1.y = -h.y; t2.x = -h.x; t2.y = h.y; ans = min(ans, ret - dis(p[b], h) + dis(p[b], t1) - dis(h, p[a]) + dis(t2, p[a])); // bh->x ab->y t1.x = p[b].x; t1.y = -p[b].y; t2.x = -p[b].x; t2.y = p[b].y; ans = min(ans, ret - dis(h, p[b]) + dis(t1, h) - dis(p[b], p[a]) + dis(t2, p[a])); ret = ans; } else if (cx == 1 && cy == 0) { Point tmp; // tmp is x axis mirror of a tmp.x = p[a].x; tmp.y = -p[a].y; ha = ret - dis(h, p[a]) + dis(h, tmp); ab = ret - dis(p[a], p[b]) + dis(tmp, p[b]); // tmp is x axis mirror of b tmp.x = p[b].x; tmp.y = -p[b].y; bh = ret - dis(h, p[b]) + dis(h, tmp); ret = min(ha, min(ab, bh)); } else if (cx == 0 && cy == 1) { Point tmp; // tmp is y axis mirror of a tmp.x = -p[a].x; tmp.y = p[a].y; ha = ret - dis(h, p[a]) + dis(h, tmp); ab = ret - dis(p[a], p[b]) + dis(tmp, p[b]); // tmp is y axis mirror of b tmp.x = -p[b].x; tmp.y = p[b].y; bh = ret - dis(h, p[b]) + dis(h, tmp); ret = min(ha, min(ab, bh)); } return ret; } int main() { // ori o.x = 0; o.y = 0; int t; cin >> t; while (cin >> p[0].x >> p[0].y >> p[1].x >> p[1].y >> h.x >> h.y) { printf("%.2f\n", calc(0, 1)); } return 0; }
HDU 3272 - Mission Impossible(计算几何),布布扣,bubuko.com
HDU 3272 - Mission Impossible(计算几何)
标签:acm
原文地址:http://blog.csdn.net/hcbbt/article/details/38614085