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…

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…

题目大意:给定一个有 N 个点,M 条边的无向图,点有点权,删除一个点就要付出所有与之有联系且没有被删除的点的点权之和的代价,求将所有点删除的最小代价是多少. 题解:从图连通性的角度出发,删除所有点就意味着需要删除所有的边.现在来考虑每条边对答案的贡献,由于所有边均需要被删除,才能使得原图完全不连通,因此只需要在删除该边的同时,将连接该边端点的两个点中点权较小的值加入答案贡献即可,即:意味着优先删除点权大的点. 代码如下 #include <bits/stdc++.h> using names…

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…