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…

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<…

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…

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…

\(>Codeforces \space 555 B. Case of Fugitive<\) 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…

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…

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…

题目:http://codeforces.com/gym/100338/attachments 贪心,每次枚举10的i次幂,除k后取余数r在用k-r补在10的幂上作为候选答案. #include<bits/stdc++.h> using namespace std; typedef unsigned long long ull; ; ull base[maxbit], n, k; void preDeal() { ] = ; ; i < maxbit; i++){ *]; } } voi…

[Codeforces 1214A]Optimal Currency Exchange(贪心) 题面 题面较长,略 分析 这个A题稍微有点思维难度,比赛的时候被孙了一下 贪心的思路是,我们换面值越小的货币越优.如有1,2,5,10,20,50,那么我们尽量用面值为1的.如果我们把原始货币换成面值为x的货币,设汇率为d,那么需要的原始货币为dx的倍数.显然dx越小,剩下的钱,即n取模dx会尽量小. 然后就可以枚举换某一种货币的数量,时间复杂度\(O(\frac{n}{d})\) 代码 #inclu…

[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…

A. Straight «A» time limit per test:1 second memory limit per test:256 megabytes input:standard input output:standard output Noora is a student of one famous high school. It's her final year in school — she is going to study in university next year.…

题目链接 Paint Tree 给你一棵n个点的树和n个直角坐标系上的点,现在要把树上的n个点映射到直角坐标系的n个点中,要求是除了在顶点处不能有线段的相交. 我们先选一个在直角坐标系中的最左下角的点,把根结点放到这个点中,然后对剩下的点进行极角排序,按逆时顺序一个个塞进来,类似地递归处理. 这样就满足了题意. #include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (int i(a); i <= (b…

Problem C. ICPC GiveawaysTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100500/attachments Description During the preparation for the ICPC contest, the organizers prepare bags full of giveaways for the contestants. Each bag us…

Balala Power! Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 2668    Accepted Submission(s): 562 Problem Description Sample Input 1 a 2 aa bb 3 a ba abc Sample Output Case #1: 25 Case #2: 132…

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…

题目:Radar Installation 对于x轴上方的每个建筑 可以计算出x轴上一段区间可以包含这个点 所以就转化成 有多少个区间可以涵盖这所有的点 排序之后贪心一下就ok 用cin 好像一直t看了好多blog 改了scanf 过了 #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> #include<cmat…

Mike has always been thinking about the harshness of social inequality. He's so obsessed with it that sometimes it even affects him while solving problems. At the moment, Mike has two sequences of positive integers A = [a1, a2, ..., an] and B = [b1, …

Physical Examination Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6155 Accepted Submission(s): 1754 Problem Description WANGPENG is a freshman. He is requested to have a physical examination wh…

Balala Power! Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 1411    Accepted Submission(s): 239 Problem Description Talented Mr.Tang has n strings consisting of only lower case characters.…

Description Fat brother and Maze are playing a kind of special (hentai) game in the playground. (Maybe it’s the OOXX game which decrypted in the last problem, who knows.) But as they don’t like using repellent while playing this kind of special (hent…

B - Sorted Adjacent Differences(CodeForces - 1339B) 题目链接 算法 思维+贪心 时间复杂度O(nlogn) 1.这道题的题意主要就是让你对一个数组进行一种特殊的排序,使得数组中相邻的两个数的差的绝对值成非递减趋势: 2.刚开始对这道题总是执拗于两个相等的数在不同位置,如何把它们放到前面这个问题,因为路走歪了,最终无果,没有思路.后来看了一些关于这道题的解题博客,豁然开朗. 3.使得数组中相邻的两个数的差的绝对值成非递减趋势,怎么想呢.单纯想怎么…

1.codeforces 349B    Color the Fence 2.链接:http://codeforces.com/problemset/problem/349/B 3.总结: 刷栅栏.1-9每个字母分别要ai升油漆,问最多可画多大的数字. 贪心,也有点考思维. #include<bits/stdc++.h> using namespace std; #define LL long long #define INF 0x3f3f3f3f int main() { ]; while(…

传送门 D. Restructuring Company time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Even the most successful company can go through a crisis period when you have to make a hard decision — to rest…

题意:给出城市(0,0),给出n个坐标,起始人数s,每个坐标k个人, 每个坐标可以覆盖到半径为r的区域,r=sqrt(x*x+y*y)的区域,问最小的半径是多少,使得城市的总人数大于等于1000000 最开始是排序,贪心来做的,发现sqrt的精度老达不到要求,于是翻了代码 于是发现用map就可以解决了 map<int,int>,it->first是第一个int的内容,it->second是第二个int的内容 话说本来是按照标签来找的,想做二分查找的题目的= = #include&l…

BerSU Ball 题目链接: http://acm.hust.edu.cn/vjudge/contest/121332#problem/E Description The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! n boys and m girls are already busy rehearsing waltz, minuet, po…

E - bits-EqualizerTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87794#problem/K Description You are given two non-empty strings S and T of equal lengths. S contains the characters 0, 1 and ?, where…

这个题目用贪心来做,关键是怎么贪心最小,那就是排序的问题了. 加入给定两个数a1, b1, a2, b2.那么如果先选1再选2的话,总的耗费就是a1 + a1 * b2 + a2; 如果先选2再选1,总的耗费就是a2 + a2 * b1 + a1.这时比较两个数的大小,发现两边都有a1+a2,所以只是比较a1*b2和a2 * b1的大小. #include <cstdio> #include <cstring> #include <algorithm> using na…

ORACLE sql 排序 根据两个条件排序,根据id号由小到大排序,同时country字段是北京的排最前面前面,其次上海,..大连,最后是其他城市,怎么写? 写法如下:select * from proviceorder by (case when country='北京' then 0 when country='上海' then 1 when country='广东' then 2 when country='深圳' then 3 when country='杭州' then 4 when…

题目链接:http://codeforces.com/gym/100269/attachments 题意: 有长度为n个格子,你有两种操作,1是放一个长度为1的东西上去,2是放一个长度为2的东西上去 每个东西在每秒钟都会产生1的能力. 然后问你怎么放才能使得最后能力最大,输出出来 解法: 贪心,最后肯定1越多越好 所以我们放1的时候,注意一下,如果放不下的话,就把其中一个2扔掉,然后放1就好了 //CF gym 100269E #include <bits/stdc++.h> using na…