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…
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…
任意门:http://codeforces.com/contest/1118/problem/F1 F1. Tree Cutting (Easy Version) time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given an undirected tree of nn vertices. Some vert…
https://codeforces.com/contest/1118/problem/F1 #include<bits/stdc++.h> using namespace std; int n; vector<int> color; vector<vector<int> > tree; ,blue=; ; pair<){ ); ); ;i<tree[v].size();i++){ int u=tree[v][i]; if(u!=p){//避免回…
题意:有\(n\)本书,A和B都至少要从喜欢的书里面读\(k\)本书,如果一本书两人都喜欢的话,那么他们就可以一起读来节省时间,问最少多长时间两人都能够读完\(k\)本书. 题解:我们可以分\(3\)种情况来存,即: ​ 1.\(a=b=1\). 2.\(a=1,b=0\). 3.\(a=0,b=1\). 对于2和3来说,我们可以将他们排序,然后合并到一起,最后放到第1种情况中再排一次序,取前\(k\)个前缀和即可. 代码: int n,k; int t,x,y; int ans; vector…
题意:给你一组数\(a\),构造一个它的子序列\(b\),然后再求\(b_1-b2+b3-b4...\),问构造后的结果最大是多少. 题解:线性DP.我们用\(dp1[i]\)来表示在\(i\)位置,并且此时子序列的长度是奇数的情况,而\(dp2\)则是偶数情况,对于每个\(a_i\),\(dp[i]\)都可以选它或者不选,拿\(dp1[i]\)举例,如果选择\(a_i\),那么状态则可以从子序列中上一个位置转移过来,所以\(dp1[i]=dp2[i-1]+a[i]\),如果不选就是\(dp1[…
题意:有一组数,每次操作可以将某个数移到头部或者尾部,问最少操作多少次使得这组数非递减. 题解:先离散化将每个数映射为排序后所对应的位置,然后贪心,求最长连续子序列的长度,那么最少的操作次数一定为\(n-len\). 感觉不好解释,直接上图,其实就是排序后它们一定是连续的,所以我们就求一个最长的连续的,然后s剩下的数移到头部尾部,贪心的想,这样一定是最优解. 代码: #include <iostream> #include <cstdio> #include <cstring…
题意:给你两个长度为\(n\)的01串\(s\)和\(t\),可以选择\(s\)的前几位,取反然后反转,保证\(s\)总能通过不超过\(3n\)的操作得到\(t\),输出变换总数,和每次变换的位置. 题解:构造题一定要充分利用题目所给的条件,对于\(s\)中的某一位i,假如它和\(t\)中的对应位置不同,我们先对前i个字符取反反转,然后再对第一个字符取反反转(就选了一个,反不反都无所谓),在取前i个位置取反反转,这样,我们就将第i个位置变换了,消耗了3次操作.这样就一定能保证在\(3n\)之内完…
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…
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> #…
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…
排序+暴力 #include<bits/stdc++.h> using namespace std; #define int long long #define N 1005000 int arr[N]; signed main(){ int _,n; cin>>_; while(_--){ cin>>n; ;i<=n;i++) cin>>arr[i]; sort(arr+,arr++n); ; ;i<=n;i++){ ; ; ;j--){ if…
题:https://codeforces.com/contest/1243/problem/D 分析:找全部可以用边权为0的点连起来的全部块 然后这些块之间相连肯定得通过边权为1的边进行连接 所以答案就是这些块的总数-1: #include<bits/stdc++.h> using namespace std; typedef long long ll; #define pb push_back ; set<int>s,g[M]; int vis[M]; void bfs(int…
这题写起来真的有点麻烦,按照官方题解的写法 先建图,然后求强连通分量,然后判断掉不符合条件的换 最后做dp转移即可 虽然看起来复杂度很高,但是n只有15,所以问题不大 #include <iostream> #include <fstream> #include <vector> #include <set> #include <map> #include <bitset> #include <algorithm> #in…
C. Sum Balance Ujan has a lot of numbers in his boxes. He likes order and balance, so he decided to reorder the numbers. There are…
D. 0-1 MST 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 description of a graph. It is an undirected weig…
C. Tile Painting Ujan has been lazy lately, but now has decided to bring his yard to good shape. First, he decided to paint the path from his house to the gate. The path consists of…
A. Maximum Square Ujan decided to make a new wooden roof for the house. He has…
Ujan has been lazy lately, but now has decided to bring his yard to good shape. First, he decided to paint the path from his house to the gate. The path consists of nn consecutive tiles, numbered from 11 to nn. Ujan will paint each tile in some color…
Ujan decided to make a new wooden roof for the house. He has nn rectangular planks numbered from 11 to nn. The ii-th plank has size ai×1ai×1 (that is, the width is 11 and the height is aiai). Now, Ujan wants to make a square roof. He will first choos…
题意:就是给你一个n,然后如果  n mod | i - j | == 0  并且 | i - j |>1 的话,那么i 和 j 就是同一种颜色,问你最大有多少种颜色? 思路: 比赛的时候,看到直接手推,发现有点东西,直接打表找出了规律 —— 如果 n的质因子只有一个,那么总数就是 那个 质因子.其它都为 1. 今天上课的时候无聊,还是试着推了一下原理. 1.如果一个数只有一个质因子 x ,那么  n-x .n-2x.n-3x ……等等全为一种颜色,也就是说每隔 x个就是同种颜色,这样的话就是有…
传送门:http://codeforces.com/contest/1092/problem/D2 D2. Great Vova Wall (Version 2) time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Vova's family is building the Great Vova Wall (named by Vo…
传送门:http://codeforces.com/contest/1092/problem/D1 D1. Great Vova Wall (Version 1) time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Vova's family is building the Great Vova Wall (named by Vo…
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…
题目链接:https://codeforces.com/contest/1203/problem/D2 题意: 给你S串.T串,问你最长删除多长的子串使得S串里仍然有T的子序列. 思路: 想了好久,先正着跑一下S串,记录T串每一个字符最左边在哪里,再倒着跑一下,记录T串的每一个字符最右边在哪里. 最后跑一下答案: 1. 开头和结尾特判一下,但不是max( L[1]-1 , l1-R[l2] ) , 而是对两个max( L[1]-1 , l1-L[l2]-1 ).max( R[1]-1 , l1-…
A. Appleman and Easy Task time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you…
一 题面 C2. Increasing Subsequence (hard version) 二 分析 需要思考清楚再写的一个题目,不能一看题目就上手,容易写错. 分以下几种情况: 1 左右两端数都小于等于构造的数组的最后一个数字 2 左右两端数至少有一个大于构造的数组最后一个数字 a. 左右两端数字相等,肯定满足上面条件.那么只可能走一个方向,两边都模拟一下,比一下大小即可 b. 左右两端数组不等,则优先看谁满足上面2的情况,若都满足则选最小的 三 AC代码 #include <bits/st…
D2. RGB Substring (hard version) inputstandard input outputstandard output The only difference between easy and hard versions is the size of the input. You are given a string…