题意:给出笛卡尔坐标系上 n 个点,n不大于100,求出这些点中能围出的最小面积. 可以肯定的是三个点围成的面积是最小的,然后就暴力枚举,计算任意三点围成的面积.刚开始是求出三边的长,然后求面积,运算步骤多,超时鸟~~,后来看了别人的代码,计算步骤挺少啊,不过我不会推这个式子. #include<stdio.h> #include<string.h> #include<math.h> struct node{ double x,y; }; node point[]; ;…

Herding Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1125    Accepted Submission(s): 325 Problem Description Little John is herding his father's cattles. As a lazy boy, he cannot tolerate cha…

题意:给出N个点的坐标,从中取些点来组成一个多边形,求这个多边形的最小面积,组不成多边形的输出"Impossible"(测试组数 T <= 25, 1 <= N <= 100,  -1000 <= 坐标Xi, Yi <= 1000). 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4709 ——>>面积最小,若有的话,一定是三角形.判断3点是否能组成一个三角形,若用斜率来做,麻烦且可能会有精度误差…

Herding Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 415    Accepted Submission(s): 59 Problem Description Little John is herding his father's cattles. As a lazy boy, he cannot tolerate chasi…

求全部点组成的三角形最小的面积,0除外. 本题就枚举全部能够组成的三角形,然后保存最小的就是答案了.由于数据量非常少. 复习一下怎样求三角形面积.最简便的方法就是向量叉乘的知识了. 并且是二维向量叉乘P1(ax, ay), P2(bx, by).公式为:|P1 X P2| = abs(ax*by - ay*bx) 三角形面积就是|P1 X P2| / 2; 本题也是float过不了.换成double就能够过了. const int MAX_N = 101; struct VertexPoint…

经过杭师大校赛的打击,明白了数学知识的重要性 开始学习数论,开始找题练手 Herding HDU - 4709 Little John is herding his father's cattles. As a lazy boy, he cannot tolerate chasing the cattles all the time to avoid unnecessary omission. Luckily, he notice that there were N trees in the m…

Herding Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2418    Accepted Submission(s): 705 Problem Description Little John is herding his father's cattles. As a lazy boy, he cannot tolerate ch…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4714 Tree2cycle Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)Total Submission(s): 400    Accepted Submission(s): 78 Problem Description A tree with N nodes and N-…

题意:给出n个点的坐标,问取出其中任意点围成的区域的最小值! 很明显,找到一个合适的三角形即可. #include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<cstdlib> #include<algorithm> using namespace std; const int maxn = 100 + 10; const int INF…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4816 2013长春区域赛的D题. 很简单的几何题,就是给了一条折线. 然后一个矩形窗去截取一部分,求最大面积. 现场跪在这题,最后时刻TLE到死,用的每一小段去三分,时间复杂度是O(n log n) , 感觉数据也不至于超时. 卧槽!!!!代码拷回来,今天在HDU一交,一模一样的代码AC了,加输入外挂6s多,不加也8s多,都可AC,呵呵·····(估计HDU时限放宽了!!!) 现场赛卡三分太SXBK…