A - Building Fence Time Limit:1000MS     Memory Limit:65535KB     64bit IO Format:%I64d & %I64u Submit Status Description Long long ago, there is a famous farmer named John. He owns a big farm and many cows. There are two kinds of cows on his farm, o…

1.三角形的所有端点 2.过所有三角形的端点对所有圆做切线,得到所有切点. 3.做任意两圆的外公切线,得到所有切点. 对上述所有点求凸包,标记每个点是三角形上的点还是某个圆上的点. 求完凸包后,因为所有点都是按逆时针(或顺时针)排好序的,如果相邻两点在同一圆上,那么求这段圆弧的距离,否则求这段直线的距离.最后得到所有周长. #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath…

Building Fence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 171    Accepted Submission(s): 25Special Judge Problem Description Long long ago, there is a famous farmer named John. He owns a bi…

题意: 给n个圆和m个三角形,且保证互不相交,用一个篱笆把他们围起来,求最短的周长是多少. 做法:--水过... 把一个圆均匀的切割成500个点,然后求凸包. 注意:求完凸包,在求周长的时候记得要把圆的那部分特殊求. sin(x)  x=PI*(i/j)://度数表示的时候用PI: #include<stdio.h> #include<iostream> #include<math.h> #include<algorithm> #define PI 3.14…

//大白p263 #include <cmath> #include <cstdio> #include <cstring> #include <string> #include <queue> #include <functional> #include <set> #include <iostream> #include <vector> #include <algorithm> u…

题目: Wall Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 119 Accepted Submission(s): 47   Problem Description Once upon a time there was a greedy King who ordered his chief Architect to build a wa…

题意:给n个圆和m个三角形,且保证互不相交,用一个篱笆把他们围起来,求最短的周长是多少. 解法1:在每个圆上均匀的取2000个点,求凸包周长就可以水过. 解法2:求出所有圆之间的外公切线的切点,以及过三角形每个顶点的的直线和圆的切点,和三角形的三个顶点.这些点做凸包确定篱笆边上的图形.凸包的边和圆弧之和即为所求.求圆弧长度的时候要判断是优弧还是劣弧.用叉积判断两个向量的方向关系即可. //Time:218MS //Memory:860K include <cstdio> #include &l…

题目链接:https://vjudge.net/problem/POJ-1873 题意:n个点(2<=n<=15),给出n个点的坐标(x,y).价值v.做篱笆时的长度l,求选择哪些点来做篱笆围住另一些点,使得选出的这些点的价值和最小,如果价值和相等要求个数最小. 思路: 看来这是WF的签到题吧.数据很小,直接二进制枚举 (1<<n),然后对未选出的点求凸包的周长,仅当选出点的长度l的和>=凸包周长时才更新答案. AC code: #include<cstdio>…

http://poj.org/problem?id=1113 题目大意:现在要给n个点,让你修一个围墙把这些点围起来,距离最小是l 分析  :现在就是求凸包的周长然后再加上一个圆的周长 #include<stdio.h> #include<string.h> #include<stdlib.h> #include<math.h> #include<algorithm> #include<iostream> #include<qu…

题目链接:https://vjudge.net/problem/POJ-1113 题意:简化下题意即求凸包的周长+2×PI×r. 思路:用graham求凸包,模板是kuangbin的. AC code: #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; ; const double PI=acos(-1.0); struct…