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…

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…

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…

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…

注意此题,每一个部分都有一个能量值v[i],他移除第i部分所需的能量是v[f[1]]+v[f[2]]+...+v[f[k]],其中f[1],f[2],...,f[k]是与i直接相连(且还未被移除)的部分的编号. 注意题目移除的都是与第i部分直接相连的部分的能量值, 将本题目简化得,只考虑两个点1和2,1和2相连,1的能量值是10,2的能量值是20, 移除绳子时,要保持能量最小,可以移除部分2,这样移除的能量就是与2相连的部分1的能量即是10: 故每次相连两部分都移除能量值大的即可 #includ…

之前一直想着建图...遍历 可是推例子都不正确 后来看数据好像看出了点规律 就抱着试一试的心态水了一下 就....过了..... 后来想想我的思路还是对的 先抽象当前仅仅有两个点相连 想要拆分耗费最小,肯定拆相应权值较小的 在这个基础上考虑问题就能够了 代码例如以下: #include <cstdio> #include <iostream> #include <algorithm> #define MAXN 10010 #define ll long long usi…

Codeforces Round #250 (Div. 2) C:http://codeforces.com/problemset/problem/437/C 题意:给以一个无向图,每个点都有一点的权值,然后如果要删除一个点的话,会有一定的费用,这个费用是与这个点的相邻的,并且是没有删除的点权值之和. 题解:很简单的,肯定是贪心,因为为了避免权值最大的点对其他点造成影响,所以首先删除的应该是权值最大值. #include<iostream> #include<cstdio> #in…

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 *******************…