地址:http://codeforces.com/contest/796/problem/D 题目: D. Police Stations time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Inzane finally found Zane with a lot of money to spare, so they togeth…

传送门 题意 n个点有n-1条边相连,其中有k个特殊点,要求: 删去尽可能多的边使得剩余的点距特殊点的距离不超过d 输出删去的边数和index 分析 比赛的时候想不清楚,看了别人的题解 一道将1个联通块转化为k个树的题目,考虑上界,应该是k-1条边,这k-1条边是原图中连接树与树的边,那么我们操作如下: 1.将特殊点i染色为i,放入队列 2.做一次bfs,对每个点相邻的点染色并判断 3.遍历边,如果边的两点不同色,则输出边的index trick 代码 #include <cstdio> #i…

C. Bank Hacking 题目大意:给出一棵n个节点的树,每个节点有一个权值,删掉一个点的代价为当前这个点的权值,并且会使其相邻点和距离为2且中间隔着未被删除的点的点权值加1,现在选一个点开始删,之后每次能删掉被删过的点的相邻点,问删掉整棵树,删各节点花费的最大值最小是多少.(n<=300,000) 思路:确定第一个删的点之后,与这个点相邻的点的删除花费是原权值加1,其他点是原权值加2,把所有点权加2后枚举一个点减2再把相邻的减1,线段树统计答案后改回去,总复杂度O(nlogn),O(n)…

A - Buying A House 题意:给你n个房间,妹子住在第m个房间,你有k块钱,你想买一个离妹子最近的房间.其中相邻的房间之间距离为10,a[i]=0表示已经被别人买了. 题解:扫一遍更新答案即可. #include<bits/stdc++.h> using namespace std; const int maxn = 105; int mp[maxn]; int n,m,k; int main(){ scanf("%d%d%d",&n,&m,&…

Description Inzane finally found Zane with a lot of money to spare, so they together decided to establish a country of their own. Ruling a country is not an easy job. Thieves and terrorists are always ready to ruin the country's peace. To fight back,…

题目链接: http://codeforces.com/problemset/problem/208/C C. Police Station time limit per test:2 secondsmemory limit per test:256 megabytes 问题描述 The Berland road network consists of n cities and of m bidirectional roads. The cities are numbered from 1 to…

题目链接:http://codeforces.com/contest/208/problem/C 思路:题目要求的是经过1~N的最短路上的某个点的路径数 /  最短路的条数的最大值.一开始我是用spfa得到从1开始的最短路和从N开始的最短路,然后分别从N开始记忆化搜索,得到从1到达最短路径上的u的路径条数,记作dp1[u], 然后再从1开始搜,得到最短路径上从N到达某个点u的路径条数,记作dp2[u],于是经过某个点u的最短路径数目为dp1[u] * dp2[u],然后只需枚举u求最大值即可.…

A. Buying A House time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output Zane the wizard had never loved anyone before, until he fell in love with a girl, whose name remains unknown to us. The gi…

题目链接:http://codeforces.com/problemset/problem/796/C 题目大意:有n家银行,第一次可以攻击任意一家银行(能量低于自身),跟被攻击银行相邻或者间接相邻(距离<=2)的银行能量+1,除了第一次外,攻击一家银行需要满足以下条件: ①:跟被攻击过后的银行相邻: ②:能量低于攻击者 ③:银行没有被攻击过 题解:可以从题意得知,比如攻击银行i,如果说银行i能量为t,跟银行距离>=2的银行中能量最大的为mx,自身至少所需能量=max(t+1,mx+2),因为…

http://codeforces.com/contest/796/problem/C Although Inzane successfully found his beloved bone, Zane, his owner, has yet to return. To search for Zane, he would need a lot of money, of which he sadly has none. To deal with the problem, he has decide…

A. Buying A House time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Zane the wizard had never loved anyone before, until he fell in love with a girl, whose name remains unknown to us. The gi…

Description Zane the wizard is going to perform a magic show shuffling the cups. There are n cups, numbered from 1 to n, placed along the x-axis on a table that has m holes on it. More precisely, cup i is on the table at the position x = i. The probl…

题目的意思就是找出未能及时处理的犯罪数, #include <iostream> using namespace std; int main(){ int n; cin >> n; , crimes = ;; ; i < n; ++ i){ cin >> a; ) recruit+=a; else recruit?recruit-- : crimes++; } cout<<crimes<<endl; }…

传送门 题意 给出n个银行,银行之间总共有n-1条边,定义i与j有边相连为neighboring,i到j,j到k有边,则定义i到k的关系为semi- neighboring, 每家银行hack的难度为a[i], 如果hack了一家银行,会使与它关系为neighboring.semi- neighboring的银行难度+1,每次hack的银行满足三个条件: 1.未被hack过 2.与hack的银行相邻,即为neighboring的关系 3.被hack的银行难度不大于 Inzane电脑的hack力(…

Description Although Inzane successfully found his beloved bone, Zane, his owner, has yet to return. To search for Zane, he would need a lot of money, of which he sadly has none. To deal with the problem, he has decided to hack the banks. There are n…

Description Zane the wizard had never loved anyone before, until he fell in love with a girl, whose name remains unknown to us. The girl lives in house m of a village. There are n houses in that village, lining in a straight line from left to right:…

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…

Codeforces Round #279 (Div. 2) 做得我都变绿了! Problems     # Name     A Team Olympiad standard input/output 1 s, 256 MB  x2377 B Queue standard input/output 2 s, 256 MB  x1250 C Hacking Cypher standard input/output 1 s, 256 MB  x740 D Chocolate standard in…

Codeforces Round #262 (Div. 2) 1003 C. Present time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Little beaver is a beginner programmer, so informatics is his favorite subject. Soon his info…

Codeforces Round #262 (Div. 2) 1004 D. Little Victor and Set time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Victor adores the sets theory. Let us remind you that a set is a group of…

A: 题目大意: 在一个multiset中要求支持3种操作: 1.增加一个数 2.删去一个数 3.给出一个01序列,问multiset中有多少这样的数,把它的十进制表示中的奇数改成1,偶数改成0后和给出的01序列相等(比较时如果长度不等各自用0补齐) 题解: 1.我的做法是用Trie数来存储,先将所有数用0补齐成长度为18位,然后就是Trie的操作了. 2.官方题解中更好的做法是,直接将每个数的十进制表示中的奇数改成1,偶数改成0,比如12345,然后把它看成二进制数10101,还原成十进制是2…

CF469 Codeforces Round #268 (Div. 2) http://codeforces.com/contest/469 开学了,时间少,水题就不写题解了,不水的题也不写这么详细了. A 水题 //#pragma comment(linker, "/STACK:102400000,102400000") #include<cstdio> #include<cmath> #include<iostream> #include<…

题目传送门 /* 贪心 + 模拟:首先,如果蜡烛的燃烧时间小于最少需要点燃的蜡烛数一定是-1(蜡烛是1秒点一支), num[g[i]]记录每个鬼访问时已点燃的蜡烛数,若不够,tmp为还需要的蜡烛数, 然后接下来的t秒需要的蜡烛都燃烧着,超过t秒,每减少一秒灭一支蜡烛,好!!! 详细解释:http://blog.csdn.net/kalilili/article/details/43412385 */ #include <cstdio> #include <algorithm> #i…

题目传送门 /* 题意:从前面找一个数字和末尾数字调换使得变成偶数且为最大 贪心:考虑两种情况:1. 有偶数且比末尾数字大(flag标记):2. 有偶数但都比末尾数字小(x位置标记) 仿照别人写的,再看自己的代码发现有清晰的思维是多重要 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath> #include <cstring> #include…

#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 #160 (Div. 1) A - Maxim and Discounts 题意 给你n个折扣,m个物品,每个折扣都可以使用无限次,每次你使用第i个折扣的时候,你必须买q[i]个东西,然后他会送你{0,1,2}个物品,但是送的物品必须比你买的最便宜的物品还便宜,问你最少花多少钱,买完m个物品 题解 显然我选择q[i]最小的去买就好了 代码 #include<bits/stdc++.h> using namespace std; const int maxn =…

Codeforces Round #383 (Div. 2) A. Arpa's hard exam and Mehrdad's naive cheat 题意 求1378^n mod 10 题解 直接快速幂 代码 #include<bits/stdc++.h> using namespace std; long long quickpow(long long m,long long n,long long k) { long long b = 1; while (n > 0) { if…