Elegant Construction                                                                         Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)                                                                          …

题目链接: Elegant Construction Time Limit: 4000/2000 MS (Java/Others)     Memory Limit: 65536/65536 K (Java/Others) Problem Description Being an ACMer requires knowledge in many fields, because problems in this contest may use physics, biology, and even…

HDU 5813 Elegant Construction(优雅建造) Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Description 题目描述 Being an ACMer requires knowledge in many fields, because problems in this contest may use physics, biology, and…

Elegant Construction 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5813 Description Being an ACMer requires knowledge in many fields, because problems in this contest may use physics, biology, and even musicology as background. And now in this prob…

Elegant Construction 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5813 Description Being an ACMer requires knowledge in many fields, because problems in this contest may use physics, biology, and even musicology as background. And now in this prob…

Being an ACMer requires knowledge in many fields, because problems in this contest may use physics, biology, and even musicology as background. And now in this problem, you are being a city architect! A city with N towns (numbered 1 through N) is und…

构造.从a[i]最小的开始放置,例如放置了a[p],那么还未放置的,还需要建边的那个点 需求量-1,然后把边连起来. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<vector> #include<…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5813 题意是:有n个点,每个点都能到达num个点,让我们构造任意一个有向图满足条件,即:使得 i 能到达 a[i] 个点: 将顶点按能到达的点数从小到大排序,排好序之后每个点只能往前面的点连边. 所以我们必须要让前面点的个数大于或等于它要连得点的个数: 因而如果存在一个排在第i位的点(前面有i-1个点),i-1>=要求到达的点数(i>要求到达的点数); 按照上述方法构造出图. 复杂度O(N^2).…

可以直接见这个博客:http://blog.csdn.net/black_miracle/article/details/52164974. 对其中的几点作一些解释: 1.这个方法我们对队列中取出的元素,把仍有出度的点连接到这个点时,这个点是不会连接到多余的点的,因为:一个点出队列的时候其他没入队的点都已经连接到这个点过了,后来再一个点出队列的时候,我们拿此时没入队的点连接到它时,肯定已经连接过之前的点了,所以肯定不会增加到达的点数.(可能我也用语言讲不太清楚233...) 2.为什么这个人的博…

6/12 2016 Multi-University Training Contest 7 期望 B Balls and Boxes(BH) 题意: n个球放到m个盒子里,xi表示第i个盒子里的球的数量,求V的期望值. 思路: 官方题解: 代码: #include <bits/stdc++.h> typedef long long ll; ll GCD(ll a, ll b) { return b ? GCD (b, a % b) : a; } int main() { ll n, m; wh…