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…

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>…

题目链接:Codeforces 437A The Child and Homework 少看了一个条件,最后被HACK掉到203名,要不然就冲到100多一点了==.. 做这个题收获最大的是英语,A twice longer than B 表示 A >= 2 * B,A twice shorter than B表示 A * 2 <= B. #include <iostream> #include <cstring> #include <cstdio> #inc…