此题的策略是选取可用范围最右边的点,一般来说该点辐射两边,左侧辐射,右侧辐射,所以用两个循环,第一个循环找出该点,第二个循环求出最右边的点 源代码: #include<iostream>#include<algorithm>using namespace std;#define maxn 1100int main(){    int r,n,k=0,num=0,x,a[maxn];    cin>>n;    for(int i=0;i<n;i++) cin>…

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 range of R units, and…

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.…

 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 : 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…

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 range of R units, and…

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…

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…

POJ3617 Best Cow Line 题意 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(或尾部)删除一个字符,加到T的尾部 目标是构造字典序尽可能小的字符串T. 思路 贪心算法,不断取S的开头和末尾中较小的一个字符放到T的末尾.但对于S的开头和末尾字符相同的情况下,需要比较下一个字符大小,这可以用如下算法实现: 按照字典序比较S和S翻转后的字符串S1,如果S较小,则从S的开头取,否则从末尾取. 代码 Source Cod…

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…