B. Balls Game Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/430/problem/B Description Iahub is training for the IOI. What is a better way to train than playing a Zuma-like game? There are n balls put in a row. Each ball i…

E. Clockwork Bomb 题目连接: http://www.codeforces.com/contest/650/problem/E Description My name is James diGriz, I'm the most clever robber and treasure hunter in the whole galaxy. There are books written about my adventures and songs about my operations…

E. Table Compression 题目连接: http://www.codeforces.com/contest/651/problem/E Description Little Petya is now fond of data compression algorithms. He has already studied gz, bz, zip algorithms and many others. Inspired by the new knowledge, Petya is now…

D. Secret Passwords One unknown hacker wants to get the admin's password of AtForces testing system, to get problems from the next contest. To achieve that, he sneaked into the administrator's office and stole a piece of paper with a list of n passwo…

暴利搜索即可 #include <iostream> #include <vector> #include <iostream> using namespace std; int main(){ int n,k,x; cin >> n >> k >> x; vector<int> c(n); ; i < n; ++ i) cin >> c[i]; ; ; i < n ; ++ i){ ]){ , r…

题意:给你n个点,m条边,然后让你使得这个这个图成为一个协和图,需要加几条边.协和图就是,如果两个点之间有一条边,那么左端点与这之间任意一个点之间都要有条边. 思路:通过并查集不断维护连通量的最大编号的节点,然后遍历即可. 代码: #include<bits/stdc++.h> using namespace std; #define int long long #define N 1005000 int f[N];int n,m; int getf(int v){// 并查集模板 if(v=…

E. Table Compression time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Little Petya is now fond of data compression algorithms. He has already studied gz, bz, zipalgorithms and many others.…

题意:对于一张图,如果$a$与$b$连通,则对于任意的$c(a<c<b)$都有$a$与$c$连通,则称该图为和谐图,现在给你一张图,问你最少添加多少条边使图变为和谐图. 思路:将一个连通块内最大的点做为根,用并查集维护,遍历一遍,对于某个点$i$及该点连通块内的根$fx$,$i$到$fx$内的每一个点,当与$i$不属于一个集合时,进行合并,答案加一,同时更新该连通块的根. #include <iostream> #include <algorithm> #include…

Little Petya is now fond of data compression algorithms. He has already studied gz, bz, zip algorithms and many others. Inspired by the new knowledge, Petya is now developing the new compression algorithm which he wants to name dis. Petya decided to…

题意:给你带边权的树,有\(m\)次询问,每次询问有多少点对\((u,v)\)之间简单路径上的最大边权不超过\(q_i\). 题解:真的想不到用最小生成树来写啊.... 我们对边权排序,然后再对询问的\(q_i\)排序,我们可以枚举\(q_i\),然后从last开始遍历边权,如果边权不大于\(q_i\),那么就可以用并查集将两个连通块合并且计数(因为我们是从小到大枚举的,所以将它们合并并不会对后面有影响,反而还会方便我们计数),\(cnt\)表示连通块的节点数,合并时贡献为\(res=cnt[f…

C. The Labyrinth 题目连接: http://www.codeforces.com/contest/616/problem/C Description You are given a rectangular field of n × m cells. Each cell is either empty or impassable (contains an obstacle). Empty cells are marked with '.', impassable cells are…

题目链接:http://codeforces.com/contest/691/problem/D 题意: 题目给出一段序列,和m条关系,你可以无限次互相交换这m条关系 ,问这条序列字典序最大可以为多少 思路: 并查集维护这m条关系,用个vector存一下当前点所能到达的点,拍下序大的优先,输出的时候输出当前点所能找到的最大的数,已输出的标记下就好了. 实现代码: #include<bits/stdc++.h> using namespace std; ; int f[M],vis[M],a[M…

C. Balls and Boxes time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Vasya had n boxes with balls in the room. The boxes stood in a row and were numbered with numbers from 1 to n from…

D. Tricky Function Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 codeforces.com/problemset/problem/429/D Description Iahub and Sorin are the best competitive programmers in their town. However, they can't both qualify to an important contest. The sele…

题目链接:http://codeforces.com/problemset/problem/429/B 给你一个矩阵,一个人从(1, 1) ->(n, m),只能向下或者向右: 一个人从(n, 1) ->(1, m),只能向上或者向右.必须有一个相遇点, 相遇点的值不能被取到, 问两个人能得到的最大路径和是多少? dp[i][j]:表示从一个点出发的最大值:先预处理从(1,1) (1,m) (n,1) (n,m)四个点出发的4个dp最大值.然后枚举所有的点,但是这个点不能在边缘,考虑枚举点不够…

题目:http://codeforces.com/problemset/problem/429/B 第一个人初始位置在(1,1),他必须走到(n,m)只能往下或者往右 第二个人初始位置在(n,1),他必须走到(1,m)只能往上或者往右 每个点都有个权值,要求两个人中间相遇一次且只有一次,相遇的那个点的权值不算,两个人的速度可以不一样,然后求出最大值是多少 思路:他的要求是相遇一次且只有一次,那么画几个图其实就只有两个情况了(这图是借了一位大佬的,[小声bb]) 既然知道了这个图分为了这四个区域,…

版权声明:本文为博主原创文章,未经博主同意不得转载. https://blog.csdn.net/yew1eb/article/details/25609981 A Points and Segments (easy)  智商题.(智商捉急~) /*********************************************************** *分析:仅仅要按Xi从小到大染成1010101010... , *1.0间隔的的序列就能保证对于随意区间[l, r]中1的个数和0的…

题目链接: http://codeforces.com/contest/429/problem/B B. Working out time limit per test2 secondsmemory limit per test256 megabytes 问题描述 Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym w…

C. Xor-tree Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/430/problem/C Description Iahub is very proud of his recent discovery, propagating trees. Right now, he invented a new tree, called xor-tree. After this new revolu…

A. Points and Segments (easy) Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/430/problem/A Description Iahub isn't well prepared on geometry problems, but he heard that this year there will be a lot of geometry problems on…

B. Working out time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Summer is coming! It's time for Iahub and Iahubina to work out, as they both want to look hot at the beach. The gym where the…

水到家了 #include <iostream> #include <vector> #include <algorithm> using namespace std; struct Point{ int index, pos; Point(, ){ index = index_; pos = pos_; } bool operator < (const Point& a) const{ return pos < a.pos; } }; int ma…

题目链接 题意: 一个n*m的矩阵,每一个方格有一个非负数,如今选择两条线路:一个左上到右下,一个左下到右上,且仅仅能有一个公共点. 求两个线路上数的最大值(公共点不算) 分析: 仅仅有两种情况,dp就可以. 记两个线路为1和2.考虑一个公共点.1为左进右出.2为下进上出.1上进下出,2为左进右出 const int MAXN = 1005; int lu[MAXN][MAXN], ld[MAXN][MAXN]; int ru[MAXN][MAXN], rd[MAXN][MAXN]; int i…

版权声明:本文为博主原创文章,未经博主同意不得转载. https://blog.csdn.net/u012476429/article/details/25607945 题目链接 题意: 给一棵树n个节点,1为根节点.操作为,选定一个节点x.当前值取反,x的孙子,孙子的孙子.. . 均取反 如今告诉初始时每一个点的值和最后每一个点的目标值.求操作次数最少时须要选择那些节点) 分析: 深度浅的点一定是受影响最小的(根节点仅仅受自己的影响).所以从根依次向下递推处理就可以 const int MAX…

题目链接 题意: n个节点,给定每一个节点的子树(包含自己)的节点个数.每一个节点假设有子节点必定大于等于2.求这种数是否存在 n (1 ≤ n ≤ 24). 分析: 用类似DP的思路,从已知開始.这题的已知显然是叶子,那么从叶子開始考虑. 如今给一个节点,子树节点数为x.那么从叶子中找x-1个就可以.之后再来一个y.不放设y <= x,这时候就有两种选择,尽量选1或者尽量选x.分析一下:首先明确一点.无论怎样选择,之后能用的点的和是一定的.假设尽量选小的,那么会使得选过之后的点数小的比較少.假…

#include <iostream> #include <string> using namespace std; int main(){ int n; cin >> n; string str; cin >> str; , x = ; ; i < n ; ++ i){ if(str[i] == 'B') cnt+=(x << i); } cout<<cnt<<endl; }   Codeforces Round…

Codeforces Round #366 (Div. 2) A I hate that I love that I hate it水题 #I hate that I love that I hate it n = int(raw_input()) s = "" a = ["I hate that ","I love that ", "I hate it","I love it"] for i in ran…

Codeforces Round #354 (Div. 2) Problems     # Name     A Nicholas and Permutation standard input/output 1 s, 256 MB    x3384 B Pyramid of Glasses standard input/output 1 s, 256 MB    x1462 C Vasya and String standard input/output 1 s, 256 MB    x1393…

直达–>Codeforces Round #368 (Div. 2) A Brain’s Photos 给你一个NxM的矩阵,一个字母代表一种颜色,如果有”C”,”M”,”Y”三种中任意一种就输出”#Color”,如果只有”G”,”B”,”W”就输出”#Black&White”. #include <cstdio> #include <cstring> using namespace std; const int maxn = 200; const int INF =…

 cf之路,1,Codeforces Round #345 (Div. 2) ps:昨天第一次参加cf比赛,比赛之前为了熟悉下cf比赛题目的难度.所以做了round#345连试试水的深浅.....       其实这个应该是昨天就写完的,不过没时间了,就留到了今天.. 地址:http://codeforces.com/contest/651/problem/A A. Joysticks time limit per test 1 second memory limit per test 256…