\(>Codeforces \space 555 B. Case of Fugitive<\) 题目大意 : 有 \(n\) 个岛屿有序排列在一条线上,第 \(i\) 个岛屿的左端点为 \(l_i\) 右端点为 \(r_i\) ,岛屿之间两两不相交, 现在对于每一个 \(1 \leq i < n\) 第 \(i\) 岛屿要和第 \(i + 1\) 岛屿之间建一座桥,桥的长度左右端点必须得在岛上.现在有 \(m\) 座已经长度建好的桥梁,试找出一种岛屿和桥匹配的方案,使得任意两座岛屿之间的…

\(>Codeforces \space 555 C. Case of Chocolate<\) 题目大意 : 有一块 \(n \times n\) 的倒三角的巧克力,有一个人要吃 \(q\) 次,每一次从一个在对角线上的点出发,向左或者向上吃若干块,直到吃到已经被吃掉的格子为止,对于每一次吃巧克力的操作,输出这一次被吃掉了多少巧克力.(文字表达不够清晰,可以配合图片理解) \(1 ≤ n ≤ 10^9\ 1 ≤ q ≤ 2 \times 10^5\) 解题思路 : 询问可以转化为找到某一行或…

B. Case of Fugitive Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/555/problem/B Description Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet almost ful…

B. Case of Fugitive time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet a…

556D - Case of Fugitive 思路:将桥长度放进二叉搜索树中(multiset),相邻两岛距离按上限排序,然后二分查找桥长度匹配并删除. 代码: #include<bits/stdc++.h> using namespace std; #define ll long long ; struct node { ll x; ll y; int id; bool operator <(const node &a)const { return y<a.y; } }…

题目连接: http://codeforces.com/problemset/problem/555/B 题目大意: 有n个岛屿(岛屿在一列上,可以看做是线性的,用来描述岛屿位置的是起点与终点),m个桥,给出每个岛屿的位置和桥的长度,问是否可以把n个岛屿连起来? 解题思路: 排序+贪心,对于n个岛屿,相邻的两个之间起点和端点可以转化为n-1个连接桥的长度区间,把区间升序排列. 对于m个桥升序排列,对于每一个桥枚举每个可以合法覆盖的区间,选取最优的,选取的时候满足L<bridge_length<…

Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet almost fully covered with water. The only dry land there is an archipelago of n narrow islands located in a row. For more comfort l…

题意:有n-1个缝隙,在上面搭桥,每个缝隙有个ll,rr值,ll<=长度<=rr的才能搭上去.求一种搭桥组合. 经典问题,应列入acm必背300题中.属于那种不可能自己想得出来的题.将二元组[ll,rr]排序(ll相同时再rr),长度x排序(升序).一个全局优先队列pq(rr小的顶部).for循环,对每个x,将ll比它小的放入优先队列pq,如果pq仍为空,说明这块桥用不上,不为空,看top的rr是否大于x,如果大于,这块桥就能用上,并且给当前的top一定是可行的. 乱码: #pragma co…

time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output Andrewid the Android is a galaxy-famous detective. He is now investigating the case of vandalism at the exhibition of contemporary art. The main…

time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output Andrewid the Android is a galaxy-famous detective. He is now investigating a case of frauds who make fake copies of the famous Stolp's gears, pu…

A. Case of the Zeros and Ones time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing…

A. Case of the Zeros and Ones time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing…

Codeforces题号:#310B 出处: Codeforces 主要算法:贪心+优先队列 难度:4.6 思路分析: 这道题乍一看没有思路…… 考虑贪心的做法.首先预处理出每两座相邻的桥之间边界相差的min和max(即题目要求的),存在b数组中.将桥的长度从小到大排序.将b数组按照min从小到大排序. 这样做有什么好处呢?我们枚举每一座桥,然后按顺序选出它适合放置的那些区间.由于这些区间都是适合放这座桥的,所以我们自然要选择差距最小的,也就是max最小的.这其实是一个贪心:让当前这座桥利用的区…

time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones. Once he thought ab…

题目大意 有一些不相交线段和一些桥,桥可以架在两个相邻的线段上.求现有的桥是否可以使所有线段连通. 题解 在两个线段上架桥,桥的长度在一个范围内,相当于一个长度的区间,一个桥只有一个长度,相当于一个长度的点.这就转化成了点匹配区间问题. 点匹配区间问题我们在贪心(POJ3614)那里学了,把所有区间按照左端点从大到小排序,把点按照位置从大到小排序,每次总是把最右侧区间与在该区间内的最右端点匹配.问题是:如何满足可以随时删除点,且可以快速找到该区间内的最右端点呢?用key值为点的长度递减的mult…

这个题跟codeforces 556 D Case of Fugitive思路一样 关于codeforces 556 D Case of Fugitive的做法的链接http://blog.csdn.net/stl112514/article/details/46868749 题意大概我方有n个军队,敌方有m个军队.军队有两个属性:攻击力和防御力 一个军队能打败还有一个军队的条件:一军队攻击力不低于还有一个军队防御力 大概是2014上海区域赛最简单的一个题? #include<map> #in…

D. Case of Fugitive time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet a…

http://codeforces.com/problemset/problem/555/B B. Case of Fugitive time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Andrewid the Android is a galaxy-famous detective. He is now chasing a cr…

Case of Fugitive CodeForces - 555B Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet almost fully covered with water. The only dry land there is an archipelago of n narrow islands l…

一.前言 今天阅读帆哥代码的时候,看到了之前没有见过的新东西, 比如java自定义注解类,如何获取注解,如何反射内部类,this$0是什么意思? 于是乎,学习并整理了一下. 二.代码示例 import java.lang.annotation.ElementType; import java.lang.annotation.Retention; import java.lang.annotation.RetentionPolicy; import java.lang.annotation.Tar…

其实是论文题.. 题意:求一个字符串中,能由单位串repeat得到的子串中,单位串重复次数最多的子串.若有多个重复次数相同的,输出字典序最小的那个. 解题思路:其实跟论文差不多,我看了很久没看懂,后来总算理解了一些.假设我们的单位串长度为l,那么我们将串划分为s[0] , s[l] , s[2*l] , s[3*l]..这样,可以根据l划分为n/l段.枚举一个j,表示当前枚举的位置为s[j*l],我们要做的是,求suf[j*l]跟suf[(j+1)*l]的lcp  (这里用rmq做,询问是o(1…

C. Case of Chocolate Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/555/problem/C Description Andrewid the Android is a galaxy-known detective. Now he does not investigate any case and is eating chocolate out of boredom. A…

C. String Manipulation 1.0 Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/555/problem/A Description Andrewid the Android is a galaxy-famous detective. He is now investigating the case of vandalism at the exhibition of cont…

B. Case of Fake Numbers Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/556/problem/B Description Andrewid the Android is a galaxy-famous detective. He is now investigating a case of frauds who make fake copies of the famou…

A. Case of the Zeros and Ones Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/556/problem/A Description Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros an…

题目传送门 /* 题意:套娃娃,可以套一个单独的娃娃,或者把最后面的娃娃取出,最后使得0-1-2-...-(n-1),问最少要几步 贪心/思维题:娃娃的状态:取出+套上(2),套上(1), 已套上(0),先从1开始找到已经套好的娃娃层数, 其他是2次操作,还要减去k-1个娃娃是只要套上就可以 详细解释:http://blog.csdn.net/firstlucker/article/details/46671251 */ #include <cstdio> #include <algor…

题目传送门 /* 题意:n个数字转盘,刚开始每个转盘指向一个数字(0~n-1,逆时针排序),然后每一次转动,奇数的+1,偶数的-1,问多少次使第i个数字转盘指向i-1 构造:先求出使第1个指向0要多少步,按照这个次数之后的能否满足要求 题目读的好累:( */ #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <cmath>…

题目传送门 /* 找规律/贪心:ans = n - 01匹配的总数,水 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <cmath> using namespace std; ; const int INF = 0x3f3f3f3f; char s[MAXN]; int main(void) //Codeforce…

[Codeforces 555E]Case of Computer Network(Tarjan求边-双连通分量+树上差分) 题面 给出一个无向图,以及q条有向路径.问是否存在一种给边定向的方案,使得这q条路径都能被满足.(如果有一条边是从a->b),而经过它的路径是从b->a,那么久不满足).只需要判断,不用输出方案. 分析 对于一个有向环,显然它可以允许各个方向的路径通过.所以我们只要把无向图里的边-双联通分量建成环,然后就不用考虑了.影响答案的只有桥. 所以我们求出所有桥,然后缩点,把图…

题目链接:http://codeforces.com/contest/556/problem/C 题目意思:有 n 个数(1,2,...,n)组成 k 条链.第 i 条链由 mi 个数组成.每一秒只可以做其中一件事:(1)将链i连到只有一个数的链 j 中(链 i 的最大值比只由单个数组成的链 j 还要小):(2)将链上最大的数从链中分离. 然后我们最终是要得到一条1,2,3,...n.问最少需要连接的秒数. 不难推出,我们需要找到编号为 1 的链包含多少个数,然后对于此链的其他数,我们需要先分离…