链接:http://poj.org/problem?id=1066 Treasure Hunt Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5431   Accepted: 2246 Description Archeologists from the Antiquities and Curios Museum (ACM) have flown to Egypt to examine the great pyramid…
题目大意:在一个正方形的迷宫里有一些交错墙,墙的两端都在迷宫的边缘墙上面,现在得知迷宫的某个位置有一个宝藏,所以需要砸开墙来获取宝藏(只能砸一段墙的中点),问最少要砸开几面墙.   分析:这个题意刚开始理解错了,以为只能砸整面墙的中点,而实际上使一段墙的中点,也就是两个交点之间的墙,这样问题就变得比较容易了,中点的意义也就不存在了,因为所有的墙都是连接的边缘,所以如果从起点到终点连线有这堵墙,那么这堵墙一定无法绕过去,所以枚举所有的边缘点到终点有几面墙即可,注意不管怎样,边缘墙一定是要砸的.  …
题意: 正方形的房子,给一些墙,墙在区域内是封闭的,给你人的坐标,每穿过一道墙需要一把钥匙,问走出正方形需要多少把钥匙. 解法: 因为墙是封闭的,所以绕路也不会减少通过的墙的个数,还不如不绕路走直线,所以枚举角度,得出直线,求出与正方形内的所有墙交点最少的值,最后加1(正方形边界). 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include…
http://poj.org/problem?id=1410 Intersection Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 11329   Accepted: 2978 Description You are to write a program that has to decide whether a given line segment intersects a given rectangle. An ex…
Treasure Hunt Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4797   Accepted: 1998 Description Archeologists from the Antiquities and Curios Museum (ACM) have flown to Egypt to examine the great pyramid of Key-Ops. Using state-of-the-ar…
Geometric Shapes Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 1243   Accepted: 524 Description While creating a customer logo, ACM uses graphical utilities to draw a picture that can later be cut into special fluorescent materials. To…
Pipe Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 8280   Accepted: 2483 Description The GX Light Pipeline Company started to prepare bent pipes for the new transgalactic light pipeline. During the design phase of the new pipe shape th…
Segments Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7739   Accepted: 2316 Description Given n segments in the two dimensional space, write a program, which determines if there exists a line such that after projecting these segments…
模板题 注意原题中说的线段其实要当成没有端点的直线.被坑了= = #include <cmath> #include <cstdio> #include <iostream> #include <cstring> using namespace std; #define eps 1e-8 #define PI acos(-1.0)//3.14159265358979323846 //判断一个数是否为0,是则返回true,否则返回false #define z…
题意:一堆线段依次放在桌子上,上面的线段会压住下面的线段,求找出没被压住的线段. sol:从下向上找,如果发现上面的线段与下面的相交,说明被压住了.break掉 其实这是个n^2的算法,但是题目已经说了没被压住的线段不超过1000个,所以不会爆 #include<math.h> #include <stdio.h> #include <string.h> ]; int n; double X1,X2,Y1,Y2; #define eps 1e-8 #define PI…