题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1392 题意:有n棵树,每棵树有一个坐标,想用一些绳子把这些树包含起来,求需要绳子的长度: 就是求凸包的周长的,把凸包各边的长度加起来就好了:注意n<=2的情况,运用GraHam算法,时间复杂度是O(nlogn); GraHam算法的过程: #include <stdio.h> #include <algorithm> #include <cstring> #inclu…

Surround the Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7164    Accepted Submission(s): 2738 Problem Description There are a lot of trees in an area. A peasant wants to buy a rope to…

Surround the Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 291    Accepted Submission(s): 140   Problem Description There are a lot of trees in an area. A peasant wants to buy a rope to…

Surround the Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 10803    Accepted Submission(s): 4187 Problem Description There are a lot of trees in an area. A peasant wants to buy a rope t…

Surround the Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9102    Accepted Submission(s): 3481 Problem Description There are a lot of trees in an area. A peasant wants to buy a rope to…

Surround the Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11811    Accepted Submission(s): 4577 Problem Description There are a lot of trees in an area. A peasant wants to buy a rope to…

链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=453 Time Limit: 2 Seconds      Memory Limit: 65536 KB 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 r…

Problem Description 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…

题意:给一堆二维的点,问你最少用多少距离能把这些点都围起来 思路: 凸包: 我们先找到所有点中最左下角的点p1,这个点绝对在凸包上.接下来对剩余点按照相对p1的角度升序排序,角度一样按距离升序排序.因为凸包有一个特点,从最左下逆时针走,所有线都在当前这条向量的左边,根据这个特点我们进行判断.我们从栈顶拿出两个点s[top-1],s[top],所以如果s[top-1] -> p[i] 在 s[top-1] -> s[top] 右边,那么s[top]就不是凸包上一点,就这样一直判断下去.判断左右可…

又是一道模板题 #include <iostream> #include <cstring> #include <cstdlib> #include <cstdio> #include <queue> #include <cmath> #include <algorithm> using namespace std; struct P { int x,y; double angle; } p[50010]; bool cm…