题目链接:http://codeforces.com/contest/731/problem/C 题解: 1.看题目时,大概知道,不同的袜子会因为要在同一天穿而差生了关联(或者叫相互制约), 其中一条袜子可以穿多次或者不穿.那么自然就想到用并查集(DSU), 把有关联的袜子放在一个集合(经过处理后,这个集合中所有的袜子的颜色一样). 2.集合问题搞定了,那么就轮到选颜色的为题了.怎么选颜色,使得每个集合的袜子颜色一样,且需要改变颜色的袜子尽可能少呢?方法是:对于每一个集合,选择袜子条数最多的那种…

Codeforces Round #582 (Div. 3)-G. Path Queries-并查集 [Problem Description] 给你一棵树,求有多少条简单路径\((u,v)\),满足\(u\)到\(v\)这条路径上的最大值不超过\(k\).\(q\)次查询. [Solution] 并查集 将所有边按权值从小到大排序,查询值\(k_i\)也从小到大排序.对于每次查询的值\(k_i\),将边权小于等于\(k_i\)的边通过并查集合并在一起.对于合并后的联通块,每个联通块对答案的贡献…

Codeforces Round #346 (Div. 2)---E. New Reform E. New Reform time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Berland has n cities connected by m bidirectional roads. No road connects a city…

C. Socks time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Arseniy is already grown-up and independent. His mother decided to leave him alone for m days and left on a vacation. She have prep…

C. Civilization Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/455/problem/C Description Andrew plays a game called "Civilization". Dima helps him. The game has n cities and m bidirectional roads. The cities are numbe…

D. Gourmet choice 链接:http://codeforces.com/contest/1131/problem/D 思路: =  的情况我们用并查集把他们扔到一个集合,然后根据 > < 跑拓扑排序,根据拓扑排序的结果从小到大填数字就好了,需要注意的细节写在代码注释里了 代码: #include<bits/stdc++.h> using namespace std; ; int f[M],n,m; set<int>st[M]; vector<int&…

#include<bits/stdc++.h>using namespace std;vector<int>g[2007];int fa[2007],vis[2007],num[2007];char s[2007][2007];int find_(int x){    if(fa[x]==x)        return x;    return fa[x]=find_(fa[x]);//合并}void dfs(int u){    vis[u]=-1;//正数说明有环,0说明需要…

E. New Reform Berland has n cities connected by m bidirectional roads. No road connects a city to itself, and each pair of cities is connected by no more than one road. It is not guaranteed that you can get from any city to any other one, using only…

ABC实在是没什么好说的,但是D题真的太妙了,详细的说一下吧 首先思路是对于a相等的分类,假设有n个,则肯定要把n-1个都增加,因为a都是相等的,所以肯定是增加t小的分类,也就是说每次都能处理一个分类,复杂度是O(n^2),这个思路很好写,优先队列随便搞一下就行了,但是题目中N = 2 * 1e5,n^2肯定是不行的,然后就是这个很巧妙的方法了,cf官方的题解我没看懂...有点尴尬,它也没给标程,但是cf的题解好像不是这个方法 看了网上用并查集的方法,首先以t为关键字从大到小排序,然后用并查集维…

种类并查集:定义种类之间的关系来判断操作是否进行 题目大意:对于题目给出的一个矩阵,我们可以进行一种操作:swap(a[i][j],a[j][i]) 使得矩阵可以变换为字典序最小的矩阵 思路: 通过扫描整个矩阵,每次都判断a[i][j] 和 a[j][i]是否需要交换 交换的前提就是: 对第i行/第j列操作,如果既对第 i 行又第 j 列进行操作等于没交换 所以我们可以将 i 和 j定义为敌人,当 他们是敌人的时候,说明需要交换 而他们是朋友的时候就说明无需交换 这里就涉及到种类并查集了,我们定…

Socks Problem Description: Arseniy is already grown-up and independent. His mother decided to leave him alone for m days and left on a vacation. She have prepared a lot of food, left some money and washed all Arseniy's clothes. Ten minutes before her…

题目链接:http://codeforces.com/problemset/problem/731/C 题意:有n只袜子,每只都有一个颜色,现在他的妈妈要去出差m天,然后让他每天穿第 L 和第 R 只袜子,他为了让每天穿的袜子都是一个颜色的,他需要把袜子涂色,共有k种颜色,求最少需要涂多少只袜子,才能保证他每天穿的袜子都是一样的. 用并查集把所有联系在一起的袜子放到一个集合中去,然后这些袜子会变成一块一块的,我们在每一块中找一种颜色最多的,然后把当前块中的其他袜子都涂成这种颜色即可,累加即可 #…

题目链接:http://codeforces.com/problemset/problem/691/D 给你n个数,各不相同,范围是1到n.然后是m行数a和b,表示下标为a的数和下标为b的数可以交换无数次.问你最后字典序最大的数列是什么. 将下面的a和b用并查集联系起来存到祖节点对应的数组中,然后从大到小排序数组,最后依次按照父节点的数组中的顺序输出. 也可以用dfs的方法解(略麻烦),形成一个环路的就在一个数组中... //并查集 #include <bits/stdc++.h> using…

D. Swaps in Permutation 题目连接: http://www.codeforces.com/contest/691/problem/D Description You are given a permutation of the numbers 1, 2, ..., n and m pairs of positions (aj, bj). At each step you can choose a pair from the given positions and swap…

题目链接:http://codeforces.com/contest/731/problem/D D. 80-th Level Archeology time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Archeologists have found a secret pass in the dungeon of one of t…

题目链接:http://codeforces.com/contest/731/problem/F 题意:有n个数,从里面选出来一个作为第一个,然后剩下的数要满足是这个数的倍数,如果不是,只能减小为他的倍数,否则就舍弃掉,然后把没有舍弃的数的值加起来,求和的最大值; 43 2 15 9 就拿这个来说,当拿3当做第一个数时结果是3+15+9=27因为2不是3的倍数:当拿2作为第一个数时,结果是2+2+14+8=26因为3,15,9都不是2的倍数,所以只能减小;同理...求最大的和; 我们可以记录每个…

A. Night at the Museum time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Grigoriy, like the hero of one famous comedy film, found a job as a night security guard at the museum. At first night…

http://codeforces.com/contest/731/problem/F 注意到一个事实,如果你要找一段区间中(从小到大的),有多少个数是能整除左端点L的,就是[L, R]这样.那么,很明显,把这个区间分段.分成[L, 2 * L - 1],因为这段区间中,都不大于等于L的两倍,这样就使得这段区间的贡献就是sum_number * L了. 就是中间有多少个数,每个数贡献一个L值.然后到枚举[2 * L, 3 * L - 1]这段区间,贡献的值就是sum_number * (2 *…

题目链接:http://codeforces.com/contest/731/problem/F F. Video Cards time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Vlad is fond of popular computer game Bota-2. Recently, the developers…

题目链接: http://codeforces.com/contest/731/problem/A A. Night at the Museum time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Grigoriy, like the hero of one famous comedy film, found a job as a…

A 模拟 #include <cstdio> #include <cstring> int Ans; ]; inline ?x:-x;} inline int Min(int x,int y) {return x>y?y:x;} int main() { ; scanf(); ;i<=strlen(Str+);i++) Ans+=Min(Abs(Str[i]-u),-Abs(Str[i]-u)),u=Str[i]; printf("%d\n",Ans…

在十五楼做的cf..一会一断...比赛的时候做出了ABCF 就没有时间了 之后没看题解写出了D..E是个神奇的博弈(递推或者dp?)看了题解也没有理解..先写了CDF.. C 有n个袜子 每个袜子都有颜色 给出m天穿的两只袜子编号 要求现在进行涂色 保证后m天每天的两只袜子都是一个颜色 可以想到和同一只袜子一起穿过的两只袜子 a == b && c == b 那就a == b 了 这样可以推出有关系的袜子都应该是一个颜色(连通分量) 由于要求出来最小涂多少只 那就bfs求一下连通分量大小…

F. Video Cards time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Vlad is fond of popular computer game Bota-2. Recently, the developers announced the new add-on named Bota-3. Of course…

题目链接:http://codeforces.com/problemset/problem/719/C C. Efim and Strange Grade time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Efim just received his grade for the last test. He studies in a…

B. War of the Corporations 题目连接: http://www.codeforces.com/contest/625/problem/B Description A long time ago, in a galaxy far far away two giant IT-corporations Pineapple and Gogol continue their fierce competition. Crucial moment is just around the…

https://codeforces.com/contest/1100/problem/F 题意 一个有n个数组c[],q次询问,每次询问一个区间的子集最大异或和 题解 单问区间子集最大异或和,线性基能处理,但是这次多次询问,假如每次重新建立基向量会超时 考虑区间的优先级,假如我只插入不删除的话,区间的优先级和左端点没有关系 贪心一下,只保留后面插入的基,这样就可以离线解决询问,然后查询的时候需要判基的位置是不是在左端点后面 代码 #include<bits/stdc++.h> #define…

B. Mahmoud and a Triangle 题目连接: http://codeforces.com/contest/766/problem/B Description Mahmoud has n line segments, the i-th of them has length ai. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn't…

E. Trains and Statistic 题目连接: http://www.codeforces.com/contest/675/problem/E Description Vasya commutes by train every day. There are n train stations in the city, and at the i-th station it's possible to buy only tickets to stations from i + 1 to a…

A. Fox and Box Accumulation 题目连接: http://codeforces.com/contest/388/problem/A Description Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top…

题目链接:http://codeforces.com/contest/613/problem/D 题意概述: 给出一棵树,每次询问一些点,计算最少删除几个点可以让询问的点两两不连通,无解输出-1.保证询问的点总数不大于300000. 分析: 先考虑遍历的做法,统计每个点代表的子树中联通询问点的数量. 这个点不是询问点:如果有至少两个不同的子树中有询问点那么当前点一定被删除,因为这个时候不删除之后这两个点就是联通的:同时除了在更深的地方遇见第一种情况之外没有必要删除那些点:没有点不用管,只有一个点…