题目大意:有个国王他有一片森林,现在他想从这个森林里面砍伐一些树木做成篱笆把剩下的树木围起来,已知每个树都有不同的价值还有高度,求出来砍掉那些树可以做成篱笆把剩余的树都围起来,要使砍伐的树木的价值最小,如果有价值相同的尽量使砍伐的树木少一些. 分析:因为树木的数量是比较少的,所以枚举所有的状态,判断那个树需要砍那个树不需要,然后按照要求求出来答案即可. 代码如下: ==================================================================…

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>…

Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 28462   Accepted: 9498 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 wo…

题目链接 题意 : 求凸包周长+一个完整的圆周长. 因为走一圈,经过拐点时,所形成的扇形的内角和是360度,故一个完整的圆. 思路 : 求出凸包来,然后加上圆的周长 #include <stdio.h> #include <string.h> #include <iostream> #include <cmath> #include <algorithm> const double PI = acos(-1.0) ; using namespac…

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…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1348 求凸包周长+2*PI*L: #include <stdio.h> #include <algorithm> #include <cstring> #include <cmath> using namespace std; ; ; ); struct point { double x, y; point(){} point(double x, double…

Surround the Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6812    Accepted Submission(s): 2594 Problem Description There are a lot of trees in an area. A peasant wants to buy a rope to…

Wall Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2848    Accepted Submission(s): 811 Problem Description Once upon a time there was a greedy King who ordered his chief Architect to build a w…

The Fortified Forest Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 6115   Accepted: 1720 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…

题目: http://poj.org/problem?id=1113 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22013#problem/F Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 26219   Accepted: 8738 Description Once upon a time there was a greedy King who…

题目链接 题意 : 让你找出最小的凸包周长 . 思路 : 用Graham求出凸包,然后对每条边求长即可. Graham详解 #include <stdio.h> #include <string.h> #include <iostream> #include <math.h> #include <algorithm> using namespace std ; struct point { int x,y ; }p[],p1[]; int n ;…

\(\color{#0066ff}{题目描述}\) 几千年前,有一个小王国位于太平洋的中部.王国的领土由两个分离的岛屿组成.由于洋流的冲击,两个岛屿的形状都变成了凸多边形.王国的国王想建立一座桥来连接这两个岛屿.为了把成本降到最低,国王要求你,主教,找到两个岛屿边界之间最小的距离. \(\color{#0066ff}{输入格式}\) 输入由几个测试用例组成. 每个测试用两个整数n,m(3≤n,m≤10000)开始 接下来的n行中的每一行都包含一对坐标,用来描述顶点在一个凸多边形中的位置. 下一条…

#include<iostream> #include<algorithm> #include<cmath> using namespace std; typedef pair<int ,int > ll; ll num,dot[1010]; int i; const double pi=3.1415926535898; ll operator -(ll a,ll b) { return make_pair(a.first-b.first,a.second-…

Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 26180   Accepted: 8081 Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bess…

BZOJ_1670_[Usaco2006 Oct]Building the Moat护城河的挖掘_求凸包 Description 为了防止口渴的食蚁兽进入他的农场,Farmer John决定在他的农场周围挖一条护城河.农场里一共有N(8<=N<=5,000)股泉水,并且,护城河总是笔直地连接在河道上的相邻的两股泉水.护城河必须能保护所有的泉水,也就是说,能包围所有的泉水.泉水一定在护城河的内部,或者恰好在河道上.当然,护城河构成一个封闭的环. 挖护城河是一项昂贵的工程,于是,节约的FJ希望护城…

题目链接:UVA 811 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 gathered by his ancestors on their travels. To protect his trees from thieves, the king or…

题目传送门 题意:砍掉一些树,用它们做成篱笆把剩余的树围起来,问最小价值 分析:数据量不大,考虑状态压缩暴力枚举,求凸包以及计算凸包长度.虽说是水题,毕竟是final,自己状压的最大情况写错了,而且忘记特判凸包点数 <= 1的情况. /************************************************ * Author :Running_Time * Created Time :2015/11/3 星期二 16:10:17 * File Name :POJ_1873…

The Fortified Forest Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 6400   Accepted: 1808 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…

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 gathered by his ancestors on their travels. To protect his trees from thieves, the king ordered that a…

n最大15,二进制枚举不会超时.枚举不被砍掉的树,然后求凸包 #include<stdio.h> #include<math.h> #include<algorithm> #include<iostream> #include <cstring> #define eps 1e-8 #define INF 1e9 using namespace std; const int MAXN = 20; struct Point { int x,y; in…

题意:是有n棵树,每棵的坐标,价值和长度已知,要砍掉若干根,用他们围住其他树,问损失价值最小的情况下又要长度足够围住其他树,砍掉哪些树.. 思路:先求要砍掉的哪些树,在求剩下的树求凸包,在判是否可行.(枚举+凸包) // Time 407ms; Memory 200K #include<iostream> #include<cstdio> #include<algorithm> #include<cmath> #include<cstring>…

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…

/* poj1873 The Fortified Forest 凸包+枚举 水题 用小树林的木头给小树林围一个围墙 每棵树都有价值 求消耗价值最低的做法,输出被砍伐的树的编号和剩余的木料 若砍伐价值相同,则取砍伐数小的方案. */ #include<stdio.h> #include<math.h> #include <algorithm> #include <vector> using namespace std; const double eps = 1…

题目链接:[http://poj.org/problem?id=1222] 题意:Light Out,给出一个5 * 6的0,1矩阵,0表示灯熄灭,反之为灯亮.输出一种方案,使得所有的等都被熄灭. 题解:首先可以用高斯消元来做,对于每个点,我们列出一个方程,左边是某个点和它相邻的点,他们的异或值等于右边的值(灯亮为1 ,灯灭为0),然后求一个异或高斯消元就可以了.可以用bitset优化,或者__int128优化(其实unsigned就可以了). 还可以枚举第一行的按开关的状态共有1<<6中状态…

LINK 题意:给出点集,每个点有个价值v和长度l,问把其中几个点取掉,用这几个点的长度能把剩下的点围住,要求剩下的点价值和最大,拿掉的点最少且剩余长度最长. 思路:1999WF中的水题.考虑到其点的数量最多只有15个,那么可以使用暴力枚举所有取点情况,二进制压缩状态,预处理出该状态下的价值,同时记录该状态拥有的点,并按价值排序.按价值枚举状态,并对拥有的这些点求凸包,check是否合法,找到一组跳出即可.然而POJ似乎没有SPJ,同样的代码POJ会超时,UVA60ms,可以在常数上优化,不预处…

题目大意: 国王有一片森林,巫师需要从所有树中选出一些做成围栏把其他树围起来, 每棵树都有其对应的价值 v 和能作为围栏的长度 l 要求最小价值,若存在多种最小价值的方案则选择余下长度更少的 树木较少 状态压缩 枚举所有状态 计算当前的状态 被选中的 树的价值和长度 其他 被围起来(未被选中)的树去求凸包 计算凸包的边长(即围栏的最小长度) 判断选中的树是否能围住凸包 再更新答案 #include <cstdio> #include <algorithm> #include <…

链接: http://poj.org/problem?id=2187 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22013#problem/E Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 24254   Accepted: 7403 Description Bessie, Farmer John's prize cow, h…

POJ 3080 Blue Jeans (求最长公共字符串) Description The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of thousands of contributors to map how the Earth was populated. As an IB…

/* poj 1474 Video Surveillance - 求多边形有没有核 */ #include <stdio.h> #include<math.h> const double eps=1e-8; const int N=103; struct point { double x,y; }dian[N]; inline bool mo_ee(double x,double y) { double ret=x-y; if(ret<0) ret=-ret; if(ret&…

Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 28157   Accepted: 9401 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 wo…