Treasure Hunt POJ - 1066
作者:互联网
Treasure Hunt
题目链接:https://vjudge.net/problem/POJ-1066
题目:
思路:将给的壁的每个端点与宝藏相连,求出当前线段与多少条线段相连,暴力求出最少相连数,瞎搞过了。。。一开始忘了当N==0情况,应当输出1
// // Created by HJYL on 2020/1/13. // #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<queue> #include<cmath> #define eps 1e-8 #define Inf 0x7fffffff //#include<bits/stdc++.h> using namespace std; const int maxn=1e5+100; struct Point{ double x,y; }; double min(double a, double b) { return a < b ? a : b; } double max(double a, double b) { return a > b ? a : b; } bool IsSegmentIntersect(Point a, Point b, Point c, Point d) { if( min(a.x, b.x) > max(c.x, d.x) || min(a.y, b.y) > max(c.y, d.y) || min(c.x, d.x) > max(a.x, b.x) || min(c.y, d.y) > max(a.y, b.y) ) return 0; double h, i, j, k; h = (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x); i = (b.x - a.x) * (d.y - a.y) - (b.y - a.y) * (d.x - a.x); j = (d.x - c.x) * (a.y - c.y) - (d.y - c.y) * (a.x - c.x); k = (d.x - c.x) * (b.y - c.y) - (d.y - c.y) * (b.x - c.x); return h * i <= eps && j * k <= eps; } int main() { int T; while(~scanf("%d",&T)) { Point a[maxn], b[maxn]; for (int i = 0; i < T; i++) scanf("%lf%lf%lf%lf", &a[i].x, &a[i].y, &b[i].x, &b[i].y); Point bao; scanf("%lf%lf", &bao.x, &bao.y); int res[maxn] = {0}; int pos = 0; for (int i = 0; i < T; i++) { for (int j = 0; j < T; j++) { if (IsSegmentIntersect(bao, a[i], a[j], b[j])) res[pos]++; } pos++; } for (int i = 0; i < T; i++) { for (int j = 0; j < T; j++) { if (IsSegmentIntersect(bao, b[i], a[j], b[j])) res[pos]++; } pos++; } sort(res, res + pos); if(T==0) printf("Number of doors = 1\n"); else printf("Number of doors = %d\n", res[0]); } return 0; }
标签:Hunt,return,min,double,Treasure,Point,POJ,max,include 来源: https://www.cnblogs.com/Vampire6/p/12193554.html