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…

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.…

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…

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…

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…

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=…

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…

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…

 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…

题目链接: http://codeforces.com/problemset/problem/650/C C. Table Compression time limit per test4 secondsmemory limit per test256 megabytes 问题描述 Little Petya is now fond of data compression algorithms. He has already studied gz, bz, zip algorithms and m…

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…

B. One Bomb time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given a description of a depot. It is a rectangular checkered field of n × m size. Each cell in a field can be empty (".&…

B. One Bomb time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given a description of a depot. It is a rectangular checkered field of n × m size. Each cell in a field can be empty (".&…

A. Joysticks time limit per test:1 second memory limit per test:256 megabytes input:standard input output:standard output Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at…

D. Zip-line 题目连接: http://www.codeforces.com/contest/650/problem/D Description Vasya has decided to build a zip-line on trees of a nearby forest. He wants the line to be as long as possible but he doesn't remember exactly the heights of all trees in t…

D. Image Preview 题目连接: http://www.codeforces.com/contest/651/problem/D Description Vasya's telephone contains n photos. Photo number 1 is currently opened on the phone. It is allowed to move left and right to the adjacent photo by swiping finger over…

C. Watchmen 题目连接: http://www.codeforces.com/contest/651/problem/C Description Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There are n watchmen on a plane, the i-th watch…

B. Beautiful Paintings 题目连接: http://www.codeforces.com/contest/651/problem/B Description There are n pictures delivered for the new exhibition. The i-th painting has beauty ai. We know that a visitor becomes happy every time he passes from a painting…

A. Joysticks 题目连接: http://www.codeforces.com/contest/651/problem/A Description Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at a1 percent and second one is charged at a2…

题目链接:http://codeforces.com/contest/699/problem/B 题解: 首先统计每行每列出现'*'的次数,以及'*'出现的总次数,得到r[n]和c[m]数组,以及sum,.然后再枚举每一个格子(O(n^2)枚举,反正输入都是枚举的). 对于每一个格子s[i][j]: 1.如果它是'*', 那么当 r[i] + c[j] - 1 = sum 时, 表明炸弹放在这个位置可以炸掉所有的墙. 2.如果它是'.', 那么当 r[i] + c[j] = sum 时, 表明炸…

A. Watchmen time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. The…

DFS A - Joysticks 嫌麻烦直接DFS暴搜吧,有坑点是当前电量<=1就不能再掉电,直接结束. #include <bits/stdc++.h> typedef long long ll; const int N = 1e5 + 5; int ans = 0; void DFS(int a, int b, int step) { if (a <= 0 || b <= 0) { ans = std::max (ans, step); return ; } if (a…

C. Table Compression 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 d…

Watchmen 题意:有n (1 ≤ n ≤ 200 000) 个点,问有多少个点的开平方距离与横纵坐标的绝对值之差的和相等: 即 = |xi - xj| + |yi - yj|.(|xi|, |yi| ≤ 109) 思路:开始想的是容斥原理,即按x,y分别排序,先计算同x的点,然后在计算同y的点,这时由于相同的点之间的连边已经算过了,这样就不能再算.并且同一个y的点中可以每个点有多个点,算是不好编码的(反正我敲了很久..WA了) 反思:上面的容斥原理是从总体的思路来考虑的,这道题的难点也就是…

Vasya's telephone contains n photos. Photo number 1 is currently opened on the phone. It is allowed to move left and right to the adjacent photo by swiping finger over the screen. If you swipe left from the first photo, you reach photo n. Similarly,…

A. Joysticks time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at…

D - Zip-line #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PII pair<int, int> using namespace std; ; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; ; ][N], stk[N],…

C. Watchmen time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. The…