HDU 4442 Physical Examination(贪心) 题目链接http://acm.split.hdu.edu.cn/showproblem.php?pid=4442 Description WANGPENG is a freshman. He is requested to have a physical examination when entering the university. Now WANGPENG arrives at the hospital. Er-.. Th…

Intelligence System Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3414    Accepted Submission(s): 1494 Problem Description After a day, ALPCs finally complete their ultimate intelligence syste…

Intelligence System Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1650    Accepted Submission(s): 722 Problem Description After a day, ALPCs finally complete their ultimate intelligence syste…

Intelligence System Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 2   Accepted Submission(s) : 1 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description After a day, ALPCs…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=3072 题目大意:为一个有向连通图加边.使得整个图全连通,有重边出现. 解题思路: 先用Tarjan把强连通分量缩点. 由于整个图肯定是连通的,所以枚举每一条边,记录到非0这个点所在连通分量的最小cost. 一共需要累加cnt-1个连通分量的cost. 在Tarjan过程中的重边,可以用链式前向星结构解决.(vector邻接表会算错) 在枚举每条边的cost中,用cost[i]记录i这个连通分量的最…

强连通分量——tarjin 算法 这道题和前面那道hdu 2767唯一不同就是,2767需要找出最小数量的边使图成为连通分量,而这个题需要一点点贪心的思想在里面,它需要求出代价最小的边使图成为连通分量: 代码: #include <cstdio> #include <cstring> #include <iostream> #include <stack> #define N 50006 using namespace std; struct Edge {…

Danganronpa 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5835 Description Chisa Yukizome works as a teacher in the school. She prepares many gifts, which consist of n kinds with a[i] quantities of each kind, for her students and wants to hold a cl…

Ball 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5821 Description ZZX has a sequence of boxes numbered 1,2,...,n. Each box can contain at most one ball. You are given the initial configuration of the balls. For 1≤i≤n, if the i-th box is empty the…

Intelligence System Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 982    Accepted Submission(s): 440 Problem Description After a day, ALPCs finally complete their ultimate intelligence system,…

题目传送门:http://acm.hdu.edu.cn/showproblem.php?pid=4004 题目意思是青蛙要过河,现在给你河的宽度,河中石头的个数(青蛙要从石头上跳过河,这些石头都是在垂直于河岸的一条直线上) 还有青蛙能够跳跃的 最多 的次数,还有每个石头离河岸的距离,问的是青蛙一步最少要跳多少米可以过河> 这是一道二分加贪心的题,从0到的河宽度开始二分,二分出一个数然后判断在这样的最小步数(一步跳多少距离)下能否过河 判断的时候要贪心 主要难在思维上,关键是要想到二分上去,能想到…