poj 1873（枚举所有的状态+凸包）

【poj 1873（枚举所有的状态+凸包）】的更多相关文章

poj 1873（枚举所有的状态+凸包）

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…

POJ 1873 The Fortified Forest（凸包）题解

题意:二维平面有一堆点,每个点有价值v和删掉这个点能得到的长度l,问你删掉最少的价值能把剩余点围起来,价值一样求删掉的点最少思路:n<=15,那么直接遍历2^15,判断每种情况.这里要优化一下,如果价值比当前最优大了continue.POJ的G++输出要用%f...orz,还是乖乖用C++... 代码: #include<set> #include<map> #include<stack> #include<cmath> #include<qu…

The Fortified Forest - POJ 1873(状态枚举+求凸包周长)

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: ==================================================================…

poj 1873 凸包+枚举

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…

●POJ 1873 The Fortified Forest

题链: http://poj.org/problem?id=1873 题解: 计算几何,凸包枚举被砍的树的集合.求出剩下点的凸包.然后判断即可. 代码: #include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; const int MAXN=16,INF=0x3f3f3f3f; co…

poj - 3254 - Corn Fields （状态压缩）

poj - 3254 - Corn Fields (状态压缩)超详细参考了 @外出散步的博客,在此基础上增加了说明题意: 农夫有一块地,被划分为m行n列大小相等的格子,其中一些格子是可以放牧的(用1标记),农夫可以在这些格子里放牛,其他格子则不能放牛(用0标记),并且要求不可以使相邻格子都有牛.现在输入数据给出这块地的大小及可否放牧的情况,求该农夫有多少种放牧方案可以选择(注意:任何格子都不放也是一种选择,不要忘记考虑! 思路: 在样例中是一个2×3的表格,我们可以用数字的二进制来枚举每一…

POJ 1873 UVA 811 The Fortified Forest (凸包 + 状态压缩枚举)

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…