City Park 题目连接: http://acm.hust.edu.cn/vjudge/contest/view.action?cid=122283#problem/F Description Porto is blessed with a beautiful city park. The park, in the western section of the city, borders the Atlantic Ocean. It has great lawns, small forest…

City Park Problem's Link: http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=129725 Mean: 在网格中给你一些矩形,求最大连通块的面积. analyse: 由于题目保证了矩形不会相交,所以不用扫描线也可做. 先把所有的线段分为横向和纵向,然后排序,依次判断是否相邻,相邻就用并查集合并,最后再用并查集统计一下面积,取最大值即可. Time complexity: O(N) Source code…

思路:逆向并查集,逆向加入每一条边即可.在获取联通块数量的时候,直接判断新加入的边是否合并了两个集合,如果合并了说明联通块会减少一个,否则不变. AC代码 #include <cstdio> #include <cmath> #include <cctype> #include <algorithm> #include <cstring> #include <utility> #include <string> #incl…

已知有 x[0-(n-1)],但是不知道具体的值,题目给定的信息 只有 I P V,说明 Xp=V,或者 I P Q V,说明 Xp ^ Xq=v,然后要求回答每个询问,询问的是 某任意的序列值 Xp1^Xp2,,,,X^pk 这个题目用加权并查集是这么处理的: 1. f[]照样是代表父节点,照样进行路径压缩,把每个 V[i]=V[i]^V[f[i]],即节点存储的值实际是它与它父亲的异或的值.为什么要这样呢,因为异或首先满足交换律,而且异或同一个数偶数次,即相当于本身,那么这个题目的其中一个要…

A secret service developed a new kind of explosive that attain its volatile property only when a specific association of products occurs. Each product is a mix of two different simple compounds, to which we call a binding pair. If N > 2, then mixing…

这题比较简单,注意路径压缩即可. AC代码 //#define LOCAL #include <stdio.h> #include <algorithm> using namespace std; const int maxn = 20000+5; int par[maxn], dis[maxn]; void init(int n) { for(int i = 0; i <= n; i++) { par[i] = i; dis[i] = 0; } } int findRoot…

Cutting Tree 题目连接: https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4922 Description Tree in graph theory refers to any connected graph (of nodes and edges) which has no simple cycle, while…

Cluster Analysis 题目连接: https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4918 Description Cluster analysis, or also known as clustering, is a task to group a set of objects into one or more…

题意:每次给出每两个数之间的大小差值.在给出关系的过程中插入询问:数a和数b的差值,若不能确定,输出UNKNOWN 解法:相对大小关系的处理:并查集 1.给出两点的相对大小关系后,找到两个点的根节点,并利用路径压缩,将两个点父亲指向根节点.然后将根节点进行合并,并给出根节点之间的相对大小关系 2.询问时,同时找到该点到根节点的距离,相减即可得到相对大小. //meek #include <iostream> #include <cstdio> #include <cmath&…

Least Crucial Node 题目链接: http://acm.hust.edu.cn/vjudge/contest/127401#problem/C Description http://7xjob4.com1.z0.glb.clouddn.com/e15d7d11650607e5795d28e1120d7109 Input There are several input lines to a test case. The first line of each test case co…