hdu 1348【凸包模板】】的更多相关文章

http://acm.hdu.edu.cn/showproblem.php?pid=1348 造城墙问题,求出凸包加上一圈圆的周长即可 凸包模板题 #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <string> #include <queue> #include <map> #include <ios…
1.HDU 1392 Surround the Trees 2.题意:就是求凸包周长 3.总结:第一次做计算几何,没办法,还是看了大牛的博客 #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm> #include<cstdio> #include<cstdlib> #define F(i,a,b) f…
Wall Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4903    Accepted Submission(s): 1419 Problem Description Once upon a time there was a greedy King who ordered his chief Architect to build a…
Wall Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3139    Accepted Submission(s): 888 Problem Description Once upon a time there was a greedy King who ordered his chief Architect to build a w…
There are a lot of trees in an area. A peasant wants to buy a rope to surround all these trees. So at first he must know the minimal required length of the rope. However, he does not know how to calculate it. Can you help him? The diameter and length…
题意:略 思路:直接套用凸包模板 #include <iostream> #include <cstdio> #include <cmath> #include <algorithm> using namespace std; #define N 50010 struct node{ int x,y,d; }p[N]; int dist(node a,node b){ return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y…
// hdu 1686 KMP模板 // 没啥好说的,KMP裸题,这里是MP模板 #include <cstdio> #include <iostream> #include <cstring> #include <algorithm> using namespace std; ; ; char T[MAX_N]; char p[MAX_M]; int f[MAX_M]; int n,m; void getfail(){ f[] = f[] = ; ;i&l…
Surround the Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10345    Accepted Submission(s): 4009 Problem Description There are a lot of trees in an area. A peasant wants to buy a rope to…
// 凸包模板 POJ1873 // n=15所以可以按位枚举求凸包,再记录数据 #include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <vector> #include <math.h> using namespace std; #define LL long long typedef pair<int,in…
给你n个点,具有速度,一个位置如果有其他点能够先到,则不能继续访问,求出里面这些点哪些点是能够无限移动的. 首先我们考虑到,一个速度小的和一个速度大的,速度小的必定只有固定他周围的一定区域是它先到的,而其他地方都是速度大的先到. 再来如果有相同速度的两点,前连线的中垂线则是它们先到的界限,如果一个点在多边形的内部,那么它必定会被与其他点连线的中垂线所包围. 因此,只要选出最大速度的点,在里面找凸包即可.但是还有很多细节,比如点重合的情况...如果速度一样的点重合,该点也不能无限移动,但是求凸包时…