CodeForces 686B-Little Robber Girl's Zoo】的更多相关文章

题意:给定一排列,让你通过一个区间交换的方式,完成排序. 析:这个题说了,最多不能超过20000次,而 n 最大才100,那么冒泡排序复杂度为 n * n,才10000,肯定是可以的,所以我们就模拟冒泡排序. 代码如下: #include <iostream> #include <cmath> #include <cstdlib> #include <set> #include <cstdio> #include <cstring>…
题目链接: B. Little Robber Girl's Zoo //#include <bits/stdc++.h> #include <vector> #include <iostream> #include <queue> #include <cmath> #include <map> #include <cstring> #include <algorithm> #include <cstdio…
B. Little Robber Girl's Zoo 题目连接: http://www.codeforces.com/contest/686/problem/B Description Little Robber Girl likes to scare animals in her zoo for fun. She decided to arrange the animals in a row in the order of non-decreasing height. However, th…
题目: 有n头数量的动物,开始它们站在一排,它们之间有高度差,所以需要将它们进行交换使得最终形成一个不减的序列,求它们交换的区间.交换的规则:一段区间[l, r]将l与l+1.l+2与l+3.....r-1与r交换. 分析: 因为n不超过100,最多的交换次数为(100-1)*(100-1)<10000(序列最坏的情况),不会超过20000:因此可以不用合并,即可以不用将区间[1,2],[3,4]合并为[1,4].采用冒泡排序就可以解决. 代码如下: #include <iostream>…
题目链接:D. The Child and Zoo 题意: 题意比较难懂,是指给出n个点并给出这些点的权值,再给出m条边.每条边的权值为该条路连接的两个区中权值较小的一个.如果两个区没有直接连接,那么f值即为从一个区走到另一个区中所经过的路中权值最小的值做为权值.如果有多条路的话,要取最大的值作为路径的长度.问,平均两个区之间移动的权值为多少. 题解: 每条边的长度已经知道了,因为路径的权值取这条路中最小的且多条路的话则取较大的,那么我们其实可以从比较大的边开始取(先把边排序),用并查集开始合并…
题意:给出一个序列,只允许进行相邻的两两交换,给出使序列变为非降序列的操作方案. 思路:关键点是操作次数不限,冒泡排序. #include<iostream> #include<cstdio> using namespace std; typedef long long ll; ll num[]; int main() { int n; while(~scanf("%d",&n)) { ;i<n;i++) { cin>>num[i];…
A题:Free Ice Cream 注意要使用LL,避免爆int #include <bits/stdc++.h> #define scan(x,y) scanf("%d%d",&x,&y) using namespace std; typedef long long LL; ; int main() { int n,has,x; while(~scan(n,has)) { LL have=has; ; ;i<n;i++) { cin>>o…
Problem_A(CodeForces 686A): 题意: \[ 有n个输入, +\space d_i代表冰淇淋数目增加d_i个, -\space d_i表示某个孩纸需要d_i个, 如果你现在手里没有\space d_i个冰淇淋, 那么这个孩纸就会失望的离开.\] 你初始有x个冰淇淋. 问最后有多少个孩纸失望的离开了. 思路: 模拟就好了, 判断当前的数目是否足够. 代码: #include <cmath> #include <cstdio> #include <cstr…
B. Little Robber Girl's Zoo time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Little Robber Girl likes to scare animals in her zoo for fun. She decided to arrange the animals in a row in the…
Spit Problem CodeForces - 29A In a Berland's zoo there is an enclosure with camels. It is known that camels like to spit. Bob watched these interesting animals for the whole day and registered in his notepad where each animal spitted. Now he wants to…
B. The Child and Zoo Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/438/problem/B Description Of course our child likes walking in a zoo. The zoo has n areas, that are numbered from 1 to n. The i-th area contains ai animal…
题目链接:Codeforces 437D The Child and Zoo 题目大意:小孩子去參观动物园,动物园分非常多个区,每一个区有若干种动物,拥有的动物种数作为该区的权值.然后有m条路,每条路的权值为该条路连接的两个区中权值较小的一个.假设两个区没有直接连接,那么f值即为从一个区走到还有一个区中所经过的路中权值最小的值做为权值.问,平均两个区之间移动的权值为多少. 解题思路:并查集+贪心.将全部的边依照权值排序,从最大的開始连接,每次连接时计算的次数为连接两块的节点数的积(乘法原理).…
Of course our child likes walking in a zoo. The zoo has n areas, that are numbered from 1 to n. The i-th area contains ai animals in it. Also there are m roads in the zoo, and each road connects two distinct areas. Naturally the zoo is connected, so…
Codeforces 437D The Child and Zoo 题目大意: 有一张连通图,每个点有对应的值.定义从p点走向q点的其中一条路径的花费为途径点的最小值.定义f(p,q)为从点p走向点q的所有路径中的最大花费.累加每一对p,q的f(p,q),并求平均值. 乍一看以为是对图的搜索,但搜索求和的过程肯定会超时.这一题巧妙的用到了并查集,因此做简单记录. 思路: 将边的权值定义为两点间的较小值,对边进行降序排序.排序后将每条边的两点进行并查集维护,由于排了序,所以可以保证两个点所属集合合…
题目链接:http://codeforces.com/problemset/problem/437/D 思路:并差集应用,先对所有的边从大到小排序,然后枚举边的时候,如果某条边的两个顶点不在同一个集合中就合并,并且用一个sum记录这两个集合的大小,这样这两个集合中的每一对点都要经过这条边,然后更新一下sum就可以了. #include <iostream> #include <cstdio> #include <cstring> #include <algorit…
D - The Child and Zoo 思路: 并查集+贪心 每条边的权值可以用min(a[u],a[v])来表示,然后按边的权值从大到小排序 然后用并查集从大的边开始合并,因为你要合并的这两个联通块之间的点肯定要经过这条边,而这条要合并的边是所有已经合并中的最小的,所以两个联通块之间的所有点之间的f就是这条边(而且是所有情况最大的,因为是从最大的边开始贪心的). 代码: #include<bits/stdc++.h> using namespace std; #define ll lon…
D. The Child and Zoo time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Of course our child likes walking in a zoo. The zoo has n areas, that are numbered from 1 to n. The i-th area contains …
time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Of course our child likes walking in a zoo. The zoo has n areas, that are numbered from 1 to n. The i-th area contains ai animals in it. Als…
/* 题意:给出n,m. n表示给出的n个横坐标为1-n,y为0的坐标m表示以下有m个坐标,在横坐标上的点 向各个角度看,在可以看到最多的点在同一条直线上的点的做多值为横坐标这一点的值,最后各个 横坐标的值的和为多少 思路:由于m的值为枚举随意的两个点连成的直线,看在直线上的点有多少,看这条线和横坐标的值为 多少.是否是整值点.假设是就记录这个整值点的最大值 */ #include<bits/stdc++.h> using namespace std; const int maxn = 1e6…
MUH and House of Cards Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 471C Description Polar bears Menshykov and Uslada from the zoo of St. Petersburg and elephant Horace from the zoo o…
C. Robbers' watch time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Robbers, who attacked the Gerda's cab, are very successful in covering from the kingdom police. To make the goal of catchi…
Polar bears Menshykov and Uslada from the zoo of St. Petersburg and elephant Horace from the zoo of Kiev got hold of lots of wooden cubes somewhere. They started making cube towers by placing the cubes one on top of the other. They defined multiple t…
A. Robbers' watch time limit per test: 2 seconds memory limit per test: 256 megabytes input: standard input output: standard output Robbers, who attacked the Gerda's cab, are very successful in covering from the kingdom police. To make the goal of ca…
题目链接: http://codeforces.com/problemset/problem/258/B B. Little Elephant and Elections time limit per test2 secondsmemory limit per test256 megabytes 问题描述 There have recently been elections in the zoo. Overall there were 7 main political parties: one…
A. Robbers' watch 题目连接: http://www.codeforces.com/contest/685/problem/A Description Robbers, who attacked the Gerda's cab, are very successful in covering from the kingdom police. To make the goal of catching them even harder, they use their own watc…
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…
D - MUH and Cube Walls Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 471D Description Polar bears Menshykov and Uslada from the zoo of St. Petersburg and elephant Horace from the zoo o…
UPD:变色了!!!历史最高1618~ Educational Codeforces Round 81 (Rated for Div. 2) The 2019 University of Jordan Collegiate Programming Contest   充实的一天,打两场可还行.补的一些题记录一下. edu81 Educational Codeforces Round 81 (Rated for Div. 2) 还不知道上分还是掉分,还挺可惜的,被B卡了没做出来,C也调了一会儿答案…
上一次我们拿学校的URP做了个小小的demo.... 其实我们还可以把每个学生的证件照爬下来做成一个证件照校花校草评比 另外也可以写一个物理实验自动选课... 但是出于多种原因,,还是绕开这些敏感话题.. 今天,我们来扒一下cf的题面! PS:本代码不是我原创 1. 必要的分析 1.1 页面的获取 一般情况CF的每一个 contest 是这样的: 对应的URL是:http://codeforces.com/contest/xxx 还有一个Complete problemset页面,它是这样的:…
The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the "root." Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that "all hou…