hdu 6184 Counting Stars(三元环计数) 题意: 给一张n个点m条边的无向图,问有多少个\(A-structure\) 其中\(A-structure\)满足\(V=(A,B,C,D)\) && \(E=(AB,BC,CD,DA,AC)\) 显然\(A-structure\)是由两个有公共边的三元环构成的 \(1 <=n <= 1e5\) \(1 <= m <= min(2e5,n*(n-1)/2)\) 思路: 三元环计数 做法1. ①统计每个点…

Problem Description Little A is an astronomy lover, and he has found that the sky was so beautiful!So he is counting stars now!There are n stars in the sky, and little A has connected them by m non-directional edges.It is guranteed that no edges conn…

Problem Description Little A is an astronomy lover, and he has found that the sky was so beautiful! So he is counting stars now! There are n stars in the sky, and little A has connected them by m non-directional edges. It is guranteed that no edges c…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6184 题意: n个点m条边的无向图,问有多少个A-structure 其中A-structure满足V=(A,B,C,D) && E=(AB,BC,CD,DA,AC) 解法: 可以看出A-structure是由两个有公共边的三元环构成的,然后就变成了这道题. http://www.cnblogs.com/spfa/p/7495438.html #include <stdio.h>…

HDU - 6184 C - Counting Stars 题目大意:有n个点,m条边,问有一共有多少个‘structure’也就是满足V=(A,B,C,D) and E=(AB,BC,CD,DA,AC)这样一个图形,类似于四边形中间连接了一条对角线. 如果我们把这个四边形拆分的话,其实就是两个共用一条边的三角形,而在图中就是三元环.求三元环有两种求法,个人感觉这个三元环的时间复杂度很玄学. 第一种一种就是枚举点x,然后枚举和点x相连的y,这时根据y的度,如果y的度小于等于sqrt(m),那么我…

题目描述 "But baby, I've been, I've been praying hard,     Said, no more counting dollars     We'll be counting stars" Grandpa Shaw loves counting stars. One evening he was sitting in his armchar, trying to figure out this problem. Consider the sky…

传送门:hdu 5862 Counting Intersections 题意:对于平行于坐标轴的n条线段,求两两相交的线段对有多少个,包括十,T型 官方题解:由于数据限制,只有竖向与横向的线段才会产生交点,所以先对横向线段按x端点排序,每次加入一个线段,将其对应的y坐标位置+1,当出现一个竖向线段时,查询它的两个y端点之间的和即为交点个数. 注意点:对x坐标排序是对所有线段端点排序:因为可能出现 “  1-1  “ 这样的情况,所以对于横着的线段,需要进行首尾x坐标处理:我的方法是对于x坐标,先…

HDU 5862 Counting Intersections(离散化+树状数组) 题目链接http://acm.split.hdu.edu.cn/showproblem.php?pid=5862 Description Given some segments which are paralleled to the coordinate axis. You need to count the number of their intersection. The input data guarant…

Counting Stars 题目链接:http://acm.xidian.edu.cn/problem.php?id=1177 离线+一维树状数组 一眼扫过去:平面区间求和,1e6的数据范围,这要hash+二维树状数组吧?这么短时间我肯定调不出来,果断弃... 结束后有人说一维树状数组可以做,ヾ(｡｀Д´｡)离线啊没想到. 然后就成了水题...(坐标要++,因为是从零开始的) 感悟:离线算法在某种情况下可以降低复杂度的维度. 代码如下: #include<cstdio> #include&l…

传送门:Hdu 5862 Counting Intersections 题意:有n条线段,每一条线段都是平行于x轴或者y轴,问有多少个交点 分析: 基本的操作流程是:先将所有的线段按照横树坐标x按小的优先排序,注意是所有的线段 :(这里是将线段都去掉只保留两个端点) 然后从左到右的顺序经行扫描,遇到横的线段,如果是左端点对应的 yi 便++ , 若是右端点对应的y1便--:  遇到竖直的线段,便统计区间[y1,y2] 的数 , 看到这了是不是有点东西了呢? 如果我是按照x排序的话,两线段若想相交…