A Bug's Life POJ - 2492 Background Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes that they feature two different genders and that they only interact with bugs of the opposite gender. In his experiment, indi…

Bug's Life Time Limit: 10000MS Memory Limit: 65536K Description Background Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes that they feature two different genders and that they only interact with bugs of the…

D - A Bug's Life 二分图 并查集 BackgroundProfessor Hopper is researching the sexual behavior of a rare species of bugs. He assumes that they feature two different genders and that they only interact with bugs of the opposite gender. In his experiment, indi…

G - A Bug's Life Time Limit:10000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2492 Appoint description: Description BackgroundProfessor Hopper is researching the sexual behavior of a rare species of bugs. H…

A Bug's Life Time Limit: 10000MS Memory Limit: 65536K Total Submissions: 28651 Accepted: 9331 Description Background Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes that they feature two different genders and…

这两个题目都是二分图的判定,用dfs染色比较容易写. 算法流程: 选取一个没有染色的点,然后将这个点染色,那么跟他相连的所有点一定是不同颜色的,所以,如果存在已经染过颜色的,如果和这个颜色相同的话,就说明不是二分图,否则,继续从这个点往下染. hihocoder #include <cstdio> #include <cstring> #include <algorithm> #include <vector> using namespace std; ;…

hdu_2444The Accomodation of Students(二分图的判定和计算) 标签:二分图匹配 题目链接 题意: 问学生是否能分成两部分,每一部分的人都不相认识,如果能分成的话,两两认识的人可以去开房哈.求最大的开房数. 题解: 二分染色,黑白染色来看这个图可以被染成一个二分图,如果可以的话求出二分图匹配的个数即可 注意事项: dfs来染色,在求解的时候注意点的编号是从0还是从1开始的 代码: //二分图最好用链表存图 #include<cstdio> #include<…

题目:戳这里 题目大意: 给你一个数列,问能否通过两个栈的push与pop把它输出成一个升序序列(每个数只能入队并出队一次) 不能的话输出0,能的话输出操作方法 主要思路: 1.判断是否可以成功输出升序序列(二分图部分) 2.能输出的话就模拟出操作方法(贪心部分) 二分图(这里只提定义,其他应用方法可见别的题解): 把一个图的所有点分成两部分,满足每一部分内部没有连边,也就是说所有的边都在这两部分之间产生. 二分图的判定: 染色法:从一个点出发,遍历临接点,把这个点直接相连的点们标记成与该点不同…

A Bug's Life Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 8528    Accepted Submission(s): 2745 Problem Description Background  Professor Hopper is researching the sexual behavior of a rare…

题目链接 题意 有一种\(bug\),所有的交往只在异性间发生.现给出所有的交往列表,问是否有可疑的\(bug\)(进行同性交往). 思路 法一:种类并查集 参考:https://www.2cto.com/kf/201310/249052.html 对于每一个集合中的元素,用一个数组\(rank\)记录它和它的祖先性别是否相同,\(0\)为相同,\(1\)为不同.每个祖先的\(rank\)值为\(0\). 路径压缩时:rank[x] = rank[prev_fa]^rank[x]; 因为是递归进…