题目大意:有一个木棒,按照顺序摆放,求出去上面没有被别的木棍压着的木棍.....   分析:可以维护一个队列,如果木棍没有被压着就入队列,如果判断被压着,就让那个压着的出队列,最后把这个木棍放进队列,不过速度并不快,枚举才是最快的......据说是任意时刻没有超过1000个top sticks.....很难注意到.   代码如下: ===================================================================================…

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…

#include<stdio.h> #include<math.h> const double eps=1e-8; int n; int cmp(double x) { if(fabs(x)<=eps)return 0; if(x<0)return -1; return 1; } struct Point { double x,y; Point (){} Point (double _x,double _y) { x=_x; y=_y; } Point operator…

Pick-up sticks Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 11884   Accepted: 4499 Description Stan has n sticks of various length. He throws them one at a time on the floor in a random way. After finishing throwing, Stan tries to fin…

Pick-up sticks Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 7699   Accepted: 2843 Description Stan has n sticks of various length. He throws them one at a time on the floor in a random way. After finishing throwing, Stan tries to find…

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…

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…

题目链接:http://poj.org/problem?id=2653 Time Limit: 3000MS Memory Limit: 65536K Description Stan has n sticks of various length. He throws them one at a time on the floor in a random way. After finishing throwing, Stan tries to find the top sticks, that…

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…

题目大意:有一个不反光并且不透光的管道,现在有一束光线从最左端进入,问能达到的最右端是多少,输出x坐标.   分析:刚开始做是直接枚举两个点然后和管道进行相交查询,不过这样做需要考虑的太多,细节不容易掌控.后来发现其实只需要对接口进行一下相交查询就简单多了,因为只需要考虑能不能通过每个截口就可以了,而且这样做的好处还有没有平行线和重叠线的情况,因为所有的截口都是垂直于x轴的,换一种想法海阔太空啊.   代码如下: =========================================…