题目链接:http://poj.org/problem?id=1228 #include<cstdio> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> #include<queue> using namespace std; ; const double PI = acos(-1.0); const double INF = 10000…

稳定凸包问题 要求每条边上至少有三个点,且对凸包上点数为1,2时要特判 巨坑无比,调了很长时间= = //POJ 1228 //稳定凸包问题,等价于每条边上至少有三个点,但对m = 1(点)和m = 2(线段)要特判 //AC 2016-10-15 #include <cstdio> #include <cstdlib> #include <iostream> #include <cmath> #define MAXN 1010 double sqr(dou…

//凸包稳定性判断:每条边上是否至少有三点 // POJ 1228 #include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <vector> #include <math.h> using namespace std; #define LL long long typedef pair<int,int> p…

Grandpa's Estate Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12337   Accepted: 3451 Description Being the only living descendant of his grandfather, Kamran the Believer inherited all of the grandpa's belongings. The most valuable one…

LINK 题意:给出一个点集,问能否够构成一个稳定凸包,即加入新点后仍然不变. 思路:对凸包的唯一性判断,对任意边判断是否存在三点及三点以上共线,如果有边不满足条件则NO,注意使用水平序,这样一来共线点的包括也较为容易,而极角序对始边和终边的共线问题较为麻烦. /** @Date : 2017-07-17 21:08:41 * @FileName: POJ 1228 稳定凸包.cpp * @Platform: Windows * @Author : Lweleth (SoungEarlf@gma…

The Fortified Forest Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 6198   Accepted: 1744 Description Once upon a time, in a faraway land, there lived a king. This king owned a small collection of rare and valuable trees, which had been…

Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 33888   Accepted: 11544 Description Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's castle. The King was so greedy, that he w…

Poj 2187 凸包模板求解 传送门 由于整个点数是50000,而求凸包后的点也不会很多,因此直接套凸包之后两重循环即可求解 #include <queue> #include <cmath> #include <cstdio> #include <cstring> #include <cstdlib> #include <iostream> #include <algorithm> #define ll long lo…

题意: 判断凸包是否稳定. 解法: 稳定凸包每条边上至少有三个点. 这题就在于求凸包的细节了,求凸包有两种算法: 1.基于水平序的Andrew算法 2.基于极角序的Graham算法 两种算法都有一个类似下面的语句: for(int i=0;i<n;i++) { while(m > 1 && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--; ch[m++] = p[i]; } 这样的话,求出来就是最简凸包,即点数尽量少的凸包,因…

找到凸包后暴力枚举边进行$check$,注意凸包是一条线(或者说两条线)的情况要输出$NO$ #include<cmath> #include<cstdio> #include<cstring> #include<algorithm> #define N 1003 #define read(x) x = getint() using namespace std; inline int getint() { int k = 0, fh = 1; char c…