B2. Character Swap (Hard Version) This problem is different from the easy version. In this version Ujan makes at most 2…

This problem is different from the easy version. In this version Ujan makes at most 2n2n swaps. In addition, k≤1000,n≤50k≤1000,n≤50 and it is necessary to print swaps themselves. You can hack this problem if you solve it. But you can hack the previou…

B1. Character Swap (Easy Version) This problem is different from the hard version. In this version Ujan makes exactly one exchange. You can hack this problem only if you solve both problems. After struggling and failing many times, Ujan decided to tr…

This problem is different from the hard version. In this version Ujan makes exactly one exchange. You can hack this problem only if you solve both problems. After struggling and failing many times, Ujan decided to try to clean up his house again. He…

链接: https://codeforces.com/contest/1234/problem/B2 题意: The only difference between easy and hard versions are constraints on n and k. You are messaging in one of the popular social networks via your smartphone. Your smartphone can show at most k most…

久违的写篇博客吧 A. Maximum Square 题目链接:https://codeforces.com/contest/1243/problem/A 题意: 给定n个栅栏,对这n个栅栏进行任意排序,问可形成的最大正方形面积是多少 分析: 水题. 先排个序 , 然后暴力枚举正方形边长就可以了 #include<bits/stdc++.h> #define ios std::ios::sync_with_stdio(false) #define sd(n) scanf("%d&qu…

Codeforces Round #599 (Div. 2) D. 0-1 MST Description Ujan has a lot of useless stuff in his drawers, a considerable part of which are his math notebooks: it is time to sort them out. This time he found an old dusty graph theory notebook with a descr…

难题不会啊…… 我感觉写这个的原因就是因为……无聊要给大家翻译题面 A. Maximum Square 简单题意: 有$n$条长为$a_i$,宽为1的木板,现在你可以随便抽几个拼在一起,然后你要从这一大块木板中裁出一块最大的正方形. $1 \leq a_i \leq n \leq 1000$ 多测,$T \leq 10$ 给个官网的图: 直接排序然后扫就行了,这数据范围是不是让你想什么神奇东西了? #include <algorithm> #include <iostream> #…

原题 https://codeforces.com/contest/1249/problem/B2 这道题一开始给的数组相当于地图的路标,我们只需对每个没走过的点进行dfs即可 #include <bits/stdc++.h> using namespace std;const int maxn=2e5+20;int a[maxn],b[maxn],c[maxn];int dfs(int pos,int step){//传递坐标与步数 if(b[pos]==1){//再次遇到b[pos],返回…

链接: http://codeforces.com/contest/1243/problem/B2 题目大意: 两个字符串,判断能否通过交换为从而使得这两个字符串完全一致,如不可以的话,直接输出NO,可以的话输出YES,并且输出每一步的交换位置. 思路:如果没个字符出现的次数为偶数次的话,那么一定可以成功,否则的话一定是NO. 如果说S[i]!=T[i],假如说,S中有与T[i]相同的元素,那么直接交换就可以了,操作次数为1,在T中找S[i]操作相同. S中没有与T[i]相同的元素,我们保证了每…