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>…

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

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: 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: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…

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…

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…

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 ;…

#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-…