1012: A MST Problem 时间限制: 1 Sec  内存限制: 32 MB 提交: 7  解决: 4 题目描述 It is just a mining spanning tree ( 最小生成树 ) problem, what makes you a little difficult is that you are in a 3D space. 输入 The first line of the input contains the number of test cases in t…

1012: A MST Problem 时间限制: 1 Sec  内存限制: 32 MB提交: 63  解决: 33[提交][状态][讨论版][命题人:外部导入] 题目描述 It is just a mining spanning tree ( 最小生成树 ) problem, what makes you a little difficult is that you are in a 3D space. 输入 The first line of the input contains the n…

u Calculate e Problem Description A simple mathematical formula for e is where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.   Output Output the approximations of e gener…

Problem A: WERTYU Description A common typing error is to place yourhands on the keyboard one row to the right of the correct position. Then ``Q'' is typed as ``W'' and ``J'' is typed as ``K'' and so on. Your task is to decode a message typed inthis…

1001: A+B Problem 时间限制: 1 Sec  内存限制: 10 MB 提交: 4864  解决: 3132 [提交][状态][讨论版] 题目描述 Calculate a+b 输入 Two integer a,b (0<=a,b<=10) 输出 Output a+b 样例输入 1 2 样例输出 3 迷失在幽谷中的鸟儿,独自飞翔在这偌大的天地间,却不知自己该飞往何方-- #include <iostream> using namespace std; int main(…

[图论01]最短路 Start Time : 2018-01-02 12:45:00    End Time : 2018-01-23 12:45:00 Contest Status : Running Current System Time : 2018-01-12 14:39:34 Solved Problem ID Title Ratio(Accepted / Submitted)   1001 最短路 51.85%(70/135)   1002 King 46.67%(35/75)  …

Codeforce 472D Design Tutorial: Inverse the Problem 解析含快速解法(MST.LCA.思維) 今天我們來看看CF472D 題目連結 題目 給你一個\(n\times n\)的矩陣代表點\(i\)到點\(j\)的最短距離.問是否可以造出一棵邊權為正的樹. 前言 這題的輸入,輸入3e6個Integer經過實測就大概需要700ms以上了(如果沒開輸入優化好像還會直接TLE的樣子).並且我一開始是用Prim's Algo去寫MST的,priority_q…

6343.Problem L. Graph Theory Homework 官方题解: 一篇写的很好的博客: HDU 6343 - Problem L. Graph Theory Homework - [(伪装成图论题的)简单数学题] 代码: //1012-6343-数学 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<bitset&g…

6330.Problem L. Visual Cube 这个题就是输出立方体.当时写完怎么都不过,后来输出b<c的情况,发现这里写挫了,判断失误.加了点东西就过了,mdzz... 代码: //1012-模拟乱写输出立方体 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<cstdlib&g…

Solution 我们考虑答案的表达式: \[ ans = \sqrt{\frac{\sum_{i = 1}^{n - 1} (w_i - \overline{w})^2}{n - 1}} \] 其中\(w[i]\)表示选择的每一条边的权值. 考虑一种朴素的做法: 我们枚举每一个\(\overline{w}\), 把所有边按照其对答案的贡献\((w_i - \overline{w})^2\)排序, 然后用普通的kruskal解决即可. 这种做法的本质在于枚举每一个可能的\((w_i - \ove…