You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12959    Accepted Submission(s): 6373 Problem Description Many geometry(几何)problems were designed in the ACM/…
You can Solve a Geometry Problem too                                         Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)                                         Problem Description Many geometry(几何)problems wer…
http://acm.hdu.edu.cn/showproblem.php?pid=1086 You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8861    Accepted Submission(s): 4317 Problem Description Many…
1016-德莱联盟 内存限制:64MB 时间限制:1000ms 特判: No通过数:9 提交数:9 难度:1 题目描述: 欢迎来到德莱联盟.... 德莱文... 德莱文在逃跑,卡兹克在追.... 我们知道德莱文的起点和终点坐标,我们也知道卡兹克的起点和 终点坐标,问:卡兹克有可能和德莱文相遇吗?,并且保证他们走的都是直线. 输入描述: 几组数据,一个整数T表示T组数据 每组数据 8个实数,分别表示德莱文的起点和终点坐标,以及卡兹克的起点和终点坐标 输出描述: 如果可能 输出 Interseeti…
You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6837 Accepted Submission(s): 3303 Problem Description Many geometry(几何)problems were designed in the ACM/ICPC. A…
You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6425    Accepted Submission(s): 3099 Problem Description Many geometry(几何)problems were designed in the ACM/I…
You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9596    Accepted Submission(s): 4725 Problem Description Many geometry(几何)problems were designed in the ACM/I…
称号: You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 145 Accepted Submission(s): 100   Problem Description Many geometry(几何)problems were designed in the ACM/IC…
You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 199    Accepted Submission(s): 132   Problem Description Many geometry(几何)problems were designed in the ACM/…
You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 13549    Accepted Submission(s): 6645 Problem Description Many geometry(几何)problems were designed in the ACM/…
You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10204    Accepted Submission(s): 5042 Problem Description Many geometry(几何)problems were designed in the ACM/…
pid=1086">You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6997    Accepted Submission(s): 3385 Problem Description Many geometry(几何)problems were designe…
You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6340    Accepted Submission(s): 3064 Problem Description Many geometry(几何)problems were designed in the ACM/I…
You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6932    Accepted Submission(s): 3350 Problem Description Many geometry(几何)problems were designed in the ACM/I…
Problem Description Many geometry(几何)problems were designed in the ACM/ICPC. And now, I also prepare a geometry problem for this final exam. According to the experience of many ACMers, geometry problems are always much trouble, but this problem is ve…
The Doors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 6734   Accepted: 2670 Description You are to find the length of the shortest path through a chamber containing obstructing walls. The chamber will always have sides at x = 0, x =…
1264 线段相交 给出平面上两条线段的两个端点,判断这两条线段是否相交(有一个公共点或有部分重合认为相交). 如果相交,输出"Yes",否则输出"No".   输入 第1行:一个数T,表示输入的测试数量(1 <= T <= 1000) 第2 - T + 1行:每行8个数,x1,y1,x2,y2,x3,y3,x4,y4.(-10^8 <= xi, yi <= 10^8) (直线1的两个端点为x1,y1 | x2, y2,直线2的两个端点为x3…
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 is these sticks such that there is no stick on top of them. Stan has noticed…
链接:传送门 题意:给出 n 个线段找到交点个数 思路:数据量小,直接暴力判断所有线段是否相交 /************************************************************************* > File Name: hdu1086.cpp > Author: WArobot > Blog: http://www.cnblogs.com/WArobot/ > Created Time: 2017年05月07日 星期日 23时34…
题目:给出一些线段,判断有几个交点. 问题:如何判断两条线段是否相交? 向量叉乘(行列式计算):向量a(x1,y1),向量b(x2,y2): 首先我们要明白一个定理:向量a×向量b(×为向量叉乘),若结果小于0,表示向量b在向量a的顺时针方向:若结果大于0,表示向量b在向量a的逆时针方向:若等于0,表示向量a与向量b平行.(顺逆时针是指两向量平移至起点相连,从某个方向旋转到另一个向量小于180度).如下图: 在上图中,OA×OB = 2 > 0, OB在OA的逆时针方向:OA×OC = -2 <…
http://acm.hdu.edu.cn/showproblem.php?pid=1086 分析:简单计算几何题,相交判断直接用模板即可. 思路:将第k条直线与前面k-1条直线进行相交判断,因为题目中不排除多条直线相交于同一个点的重复情况. 代码: #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include <queue> #i…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1086 判断两条线段是否有交点,我用的是跨立实验法: 两条线段分别是A1到B1,A2到B2,很显然,如果这两条线段有交点,那么可以肯定的是: A1-B1,A2-B1这两个向量分别在B2-B1的两边,判断是不是在两边可以用向量的叉积来判断,这里就不说了,同理B1-A1,B2-A1在A2-A1的两边,当同时满足这两个条件时,说明这两条线段是有交点的. #include<cstdio> #include&…
数学题,证明AB和CD.只需证明C.D在AB直线两侧,并且A.B在CD直线两侧.公式为:(ABxAC)*(ABxAD)<= 0 and(CDxCA)*(CDxCB)<= 0 #include <stdio.h> #define MAXNUM 105 typedef struct { double x1, y1; double x2, y2; } line_st; line_st lines[MAXNUM]; int cal(int i, int j) { double ab_x,…
Problem Description Many geometry(几何)problems were designed in the ACM/ICPC. And now, I also prepare a geometry problem for this final exam. According to the experience of many ACMers, geometry problems are always much trouble, but this problem is ve…
pid=5323" target="_blank" style="">链接 Solve this interesting problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 511    Accepted Submission(s): 114 Problem Descriptio…
题目传送门 /* 题意:告诉一个区间[L,R],问根节点的n是多少 DFS+剪枝:父亲节点有四种情况:[l, r + len],[l, r + len - 1],[l - len, r],[l - len -1,r]; */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #include <queue> using namespace std;…
转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud Solve this interesting problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1731    Accepted Submission(s): 519 Problem Description…
Solve this interesting problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1479    Accepted Submission(s): 423 Problem Description Have you learned something about segment tree? If not, don’…
直接用kuangbin的板子,能判不规范,规范和不交 另外线段在矩形内也可以,判断方式是比较线段的端点和矩形四个角 #include <cstdio> #include <cmath> #include <algorithm> #include <iostream> using namespace std; ; ; )?:-;} struct Point{ double x,y; Point(,):x(_x),y(_y){} bool operator ==…
题目大意: 要完成两种属性p,q的需求,给定n个双属性物品及其单位个物品中含有的属性,要求选择最少的物品来达成属性需求.(可以选择实数个物品) 题解: 实际上是一种属性混合问题 我们知道对于两种双属性物品,按照一定比例融合 可以配置出的物品的属性在二维平面上的分布一定是一条直线 而这条直线由最初的双属性物品所对应的点坐标所确定 扩展到三个物品,我们发现所有可配置的物品构成了一个三角形 扩展到n个物品,我们发现这n个点构成的凸包内的物品一定都可以配置 所以我们求出凸包来 然后我们从原点想我们的需求…