带来两题贪心算法的题. 1.给定长度为N的字符串S,要构造一个长度为N的字符串T.起初,T是一个空串,随后反复进行下面两个操作:1.从S的头部删除一个字符,加到T的尾部.2.从S的尾部删除一个字符,加到T的尾部.求你任意采取这两个步骤后能得到的最小字符串T 2.直线上有N个点,点i的位置是xi,这N个点中选择若干个做上标记,对于每个点,在他们距离为R的区域内必须带有标记点,求在满足这个条件的情况下,所需要标记点的最少个数. 1.POJ 3617 Best Cow Line http://poj.…

POJ 3617 Best Cow Line Time Limit: 1000MS Memory Limit: 65536K [Description] [题目描述] FJ is about to take his N (1 ≤ N ≤ 2,000) cows to the annual "Farmer of the Year" competition. In this contest every farmer arranges his cows in a line and herds…

Best Cow Line   Time Limit: 1000MS      Memory Limit: 65536K Total Submissions: 16104    Accepted: 4547 Description FJ is about to take his N (1 ≤ N ≤ 2,000) cows to the annual"Farmer of the Year" competition. In this contest every farmer arrang…

Best Cow Line Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9284   Accepted: 2826 Description FJ is about to take his N (1 ≤ N ≤ 2,000) cows to the annual"Farmer of the Year" competition. In this contest every farmer arranges his…

Best Cow Line Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 26670   Accepted: 7226 Description FJ is about to take his N (1 ≤ N ≤ 2,000) cows to the annual"Farmer of the Year" competition. In this contest every farmer arranges his…

Best Cow Line Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 42701   Accepted: 10911 Description FJ is about to take his N (1 ≤ N ≤ 2,000) cows to the annual"Farmer of the Year" competition. In this contest every farmer arranges hi…

 Saruman's Army Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 3069 Appoint description:  System Crawler  (2015-04-27) Description Saruman the White must lead his army along a straight path fro…

Saruman's Army Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18794   Accepted: 9222 Description Saruman the White must lead his army along a straight path from Isengard to Helm’s Deep. To keep track of his forces, Saruman distributes s…

http://poj.org/problem;jsessionid=F0726AFA441F19BA381A2C946BA81F07?id=3617 Description FJ is about to take his N (1 ≤ N ≤ 2,000) cows to the annual"Farmer of the Year" competition. In this contest every farmer arranges his cows in a line and her…

题目链接:http://poj.org/problem?id=3617 题目意思:给出一条长度为n的字符串S,目标是要构造一条字典序尽量小,长度为n的字符串T.构造的规则是,如果S的头部的字母 < S的尾部的字母,那么将S的头部的字母加入到T中,删除S的头部的字母:如果S的头部的字母 > S的尾部的字母,那么将S的尾部的字母加入到T中,删除S的尾部的字母. 这个题目的关键是如何处理 S 的头部的字母(假设用 i 指示) = S的尾部的字母(j) 这种情况.此时需要比较 i+1 和 j-1 的位…

原题链接:http://poj.org/problem?id=3617 问题梗概:给定长度为 的字符串 , 要构造一个长度为 的字符串 .起初, 是一个空串,随后反复进行下列任意操作. 从 的头部删除一个字符,加到 的尾部. 从 的尾部删除一个字符,加到 的尾部. 目的是要构造字典序尽可能小的字符串 .     限制条件: 字符串 只包含大写英文字母 输出的字符串每 80 个字符进行一次换行 字典序是指从前到后比较两个字符串大小的方法.首先比较第 1 个字符,如果不同则第 1 个字符较小的字符串…

题目链接 Description FJ is about to take his N (1 ≤ N ≤ 2,000) cows to the annual"Farmer of the Year" competition. In this contest every farmer arranges his cows in a line and herds them past the judges. The contest organizers adopted a new registra…

题意:给定一行字符串,让你把它变成字典序最短,方法只有两种,要么从头部拿一个字符,要么从尾部拿一个. 析:贪心,从两边拿时,哪个小先拿哪个,如果一样,接着往下比较,要么比到字符不一样,要么比完,也就是说从头部和尾部拿都一样,那么就随便拿一个了. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <…

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> using namespace std; int ma…

地址 http://poj.org/problem?id=3069 题解 题目可以考虑贪心 尽可能的根据题意选择靠右边的点 注意 开始无标记点 寻找左侧第一个没覆盖的点 再来推算既可能靠右的标记点为一轮 我最开始就是轮次的操作理解错误 结果wa了 ac代码如下 #include <iostream> #include <vector> #include <algorithm> using namespace std; /* poj3069 题目大意:一个直线上有N个点.…

POJ 3069 Saruman's Army(萨鲁曼军) Time Limit: 1000MS   Memory Limit: 65536K [Description] [题目描述] Saruman the White must lead his army along a straight path from Isengard to Helm’s Deep. To keep track of his forces, Saruman distributes seeing stones, know…

Saruman's Army Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8477   Accepted: 4317 Description Saruman the White must lead his army along a straight path from Isengard to Helm’s Deep. To keep track of his forces, Saruman distributes se…

Saruman's Army Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other) Total Submission(s) : 3   Accepted Submission(s) : 2 Problem Description Saruman the White must lead his army along a straight path from Isengard to Helm…

题目连接 Description Saruman the White must lead his army along a straight path from Isengard to Helm's Deep. To keep track of his forces, Saruman distributes seeing stones, known as palantirs, among the troops. Each palantir has a maximum effective rang…

在一条直线上,有n个点.从这n个点中选择若干个,给他们加上标记.对于每一个点,其距离为R以内的区域里必须有一个被标记的点.问至少要有多少点被加上标记 Saruman the White must lead his army along a straight path from Isengard to Helm's Deep. To keep track of his forces, Saruman distributes seeing stones, known as palantirs, am…

问题: 链接:http://poj.org/problem?id=3617 思路: 按照字典序比较S和将S反转后的字符串S' 如果S较小,就从S的开头取出一个字符,加到T的末尾(更新下标值) 如果S’较小,从S’的末尾取出一个字符,加到T的末尾(更新下标值) 代码: # include <iostream> # include <string> using namespace std; int N; ]; ]; int main() { cin>>N; ; i <…

题目: 给定长度为N的字符串S,要构造一个长度为N的字符串T.起初,T是一个空串,随后反复进行下列任意操作. ·从S的头部删除一个字符,加到T的尾部 ·从S的尾部删除一个字符,加到T的尾部 目标是要构造字典序尽可能小的字符串T. #include "iostream" #include "cstring" #include "vector" using namespace std; #define MAX_N 1000 void solve(in…

链接:http://poj.org/problem?id=3069 题解 #include<iostream> #include<algorithm> using namespace std; ; int N,R; // N是部队数,R是有效射程 int X[MAX]; void solve(){ sort(X,X+N); ,ans=; while(i<N){ // s是没有被覆盖的最左边的点的位置 int s=X[i++]; //一直向右前进直到距 s的距离大于 R的点,此…

简单贪心. 从左边开始,找 r 以内最大距离的点,再在该点的右侧找到该点能覆盖的点.如图. 自己的逻辑有些混乱,最后还是参考书上代码.(<挑战程序设计> P46) /****************************************** Problem: 3069 User: Memory: 668K Time: 16MS Language: G++ Result: Accepted ******************************************/ #inc…

给定长度为N的字符串S,要构造一个长度为N的字符串T,起初,T是一个空串,随后反复进行下列任意操作. *从S的头部删除一个字符,加到T的尾部 *从S的尾部删除一个字符,加到T的尾部 目标是要构造字典序尽可能小的字符串T 模拟情景,想出来一个有意思的比喻,写出来程序,俩个推土机开始在互相看不到对方的直线工作推物品,每个人每次走一步,并且先将物品重量较小的先推走, 由于没有手机,只能通过副驾驶通信员互相沟通,谁的前方物品较轻先推走,于是通信员下车检查,物品,当A,B通信员发现一方较轻时,就全部回各自…

题意:给你一个字符串,让你重新排列,只能从头或者尾部取出一个放到新字符串队列的最后.按照字典序. 解决方法:比较前后两个的大小,谁小输出谁,相等,就往当中比来确定当前应该拿最前面的还是最后面的,如果再相等就继续.... 所以比较这个动作的单一功能,可以写成一个check函数,方便操作也方便递归. #include<iostream> #include<cstring> using namespace std; #define MAX 30005 char str[MAX]; int…

题目不算难,但是不认真想的话很容易wa,我就是wa了多次才意识到自己想法存在的缺陷. 相同的时候往后找知道出现不相同时,只能判断出当前字符的优先顺序. 这个题目如果朴素的按照这种方法做的话复杂度其实是n*n的,可是数据较弱,可以过,我用的就是朴素的办法. 但是多加思索的话可以发现我们可以用后缀数组保存原串和反串,使复杂度降低到nlongn. #include <iostream> #include <cstdio> #include <cstring> using na…

题目链接 题目描述 对于一个给定的字符串,可以从左右两端取字符,依次排列构成一个新的字符串. 求可能构成的字符串中字典序 最小的一个. 例:ACDBCB -> ABCBCD 思路 参考自 xueyifan1993. 正确的 贪心 姿势: 记左端位置为 \(l\),右端位置为 \(r\),比较 \(suffix(l)\) 与 \(inv(prefix(r))\),取小者. 比较 则显然是借助 \(rank\) 数组. // 以及正确的I/O姿势真的重要... // scanf+printf:117…

题目传送门 1 2 题意:有向图,所有点先走到x点,在从x点返回,问其中最大的某点最短路程 分析:对图正反都跑一次最短路,开两个数组记录x到其余点的距离,这样就能求出来的最短路以及回去的最短路. POJ 3268 //#include <bits/stdc++.h> #include <cstdio> #include <queue> #include <algorithm> #include <cstring> using namespace…

P2870 [USACO07DEC]最佳牛线,黄金Best Cow Line, Goldpoj 3617 http://poj.org/problem?id=3617 题目描述FJ is about to take his N (1 ≤ N ≤ 500,000) cows to the annual”Farmer of the Year” competition. In this contest every farmer arranges his cows in a line and herds…