题目连接:Codeforces 437C  The Child and Toy 贪心,每条绳子都是须要割断的,那就先割断最大值相应的那部分周围的绳子. #include <iostream> #include <cstring> #include <cstdio> #include <algorithm> using namespace std; const int MAX_N = 1000 + 10; int G[MAX_N][MAX_N]; struct…

time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output On Children's Day, the child got a toy from Delayyy as a present. However, the child is so naughty that he can't wait to destroy the toy. The…

题目链接:Codeforces 437D The Child and Zoo 题目大意:小孩子去參观动物园,动物园分非常多个区,每一个区有若干种动物,拥有的动物种数作为该区的权值.然后有m条路,每条路的权值为该条路连接的两个区中权值较小的一个.假设两个区没有直接连接,那么f值即为从一个区走到还有一个区中所经过的路中权值最小的值做为权值.问,平均两个区之间移动的权值为多少. 解题思路:并查集+贪心.将全部的边依照权值排序,从最大的開始连接,每次连接时计算的次数为连接两块的节点数的积(乘法原理).…

意甲冠军:给定一个无向图,每个小点右键.操作被拉动所有点逐一将去,直到一个点的其余部分,在连边和点拉远了点,在该点右点的其余的费用.寻找所需要的最低成本的运营完全成本. 解法:贪心的思想,每次将剩余点中点权最大的点揪出,这样能够保证每条边都是会选择相对小的点权被消耗掉.所以直接输出全部边的边权和就可以. 代码: /****************************************************** * author:xiefubao *******************…

A. The Child and Toy Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/438/problem/A Description On Children's Day, the child got a toy from Delayyy as a present. However, the child is so naughty that he can't wait to destroy…

题目例如以下: C. The Child and Toy time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output On Children's Day, the child got a toy from Delayyy as a present. However, the child is so naughty that he can't…

The Child and Toy time limit per test1 second On Children's Day, the child got a toy from Delayyy as a present. However, the child is so naughty that he can't wait to destroy the toy. The toy consists of n parts and m ropes. Each rope links two parts…

题目链接:Codeforces 437B The Child and Set 開始是想到了这样的情况,比方lowbit之后从大到小排序后有这么几个数,200.100,60.50.S = 210.那先选了200肯定就不行了,所以写了个DFS.果断TLE.后来想了想lowbit之后的值都是最高位是1.其余位都是0的二进制数,所以上边的情况是不会出现的,至于原因我感觉我说不清楚.. #include <iostream> #include <cstdio> #include <ve…

C. The Child and Toy time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output On Children's Day, the child got a toy from Delayyy as a present. However, the child is so naughty that he can't wait to…

题目链接:Codeforces 437E The Child and Polygon 题目大意:给出一个多边形,问说有多少种切割方法.将多边形切割为多个三角形. 解题思路:首先要理解向量叉积的性质,一開始将给出的点转换成顺时针.然后用区间dp计算.dp[i][j]表示从点i到点j能够有dp[i][j]种分割方法.然后点i和点j能否够做为分割线.要经过推断.即在i和j中选择的话点k的话,点k要在i,j的逆时针方. #include <cstdio> #include <cstring>…