POJ3617 Best Cow Line
来源:互联网 发布:台风战机参数知乎 编辑:程序博客网 时间:2024/05/16 09:42
文章链接:http://poj.org/problem?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 herds them past the judges.
The contest organizers adopted a new registration scheme this year: simply register the initial letter of every cow in the order they will appear (i.e., If FJ takes Bessie, Sylvia, and Dora in that order he just registers BSD). After the registration phase ends, every group is judged in increasing lexicographic order according to the string of the initials of the cows' names.
FJ is very busy this year and has to hurry back to his farm, so he wants to be judged as early as possible. He decides to rearrange his cows, who have already lined up, before registering them.
FJ marks a location for a new line of the competing cows. He then proceeds to marshal the cows from the old line to the new one by repeatedly sending either the first or last cow in the (remainder of the) original line to the end of the new line. When he's finished, FJ takes his cows for registration in this new order.
Given the initial order of his cows, determine the least lexicographic string of initials he can make this way.
Input
* Line 1: A single integer: N
* Lines 2..N+1: Line i+1 contains a single initial ('A'..'Z') of the cow in the ith position in the original line
Output
The least lexicographic string he can make. Every line (except perhaps the last one) contains the initials of 80 cows ('A'..'Z') in the new line.
Sample Input
6ACDBCB
Sample Output
ABCBCD
Source
这是一道贪心的题,根据给的字符串,每次从字符串的“头部”或“尾部”拿掉一个组成一个新的字符串,是这个新组成的字符串的字典序尽可能的小, 所以,每次选择字典序小的字符来添加到新字符串里边,如果头部和尾部的字典序一样,那就比下一位,直到不相同为止
代码如下:
#include <iostream>#include <cstdio>#include <cstring>using namespace std;int main(){ int n; while(~scanf("%d",&n)) { string s; for(int i=0;i<n;i++) { getchar(); char temp; scanf("%c",&temp); s += temp; } int len = s.length() - 1; int Count = 0; int lengthCount = 0; while(Count<=len) { bool left = false; for(int i=0;Count+i<=len;i++) { if(s[Count+i]<s[len-i]) { left = true; break; } else if(s[Count+i]>s[len-i]) { left = false; break; } } if(left) printf("%c",s[Count++]); else printf("%c",s[len--]); lengthCount++; lengthCount%=80; if(lengthCount%80==0) printf("\n"); } if(s.length()%80) printf("\n"); } return 0;}
- POJ3617 Best Cow Line
- POJ3617 Best Cow Line
- POJ3617 Best Cow Line
- best cow line(poj3617)
- POJ3617-Best Cow Line
- POJ3617 Best Cow Line
- POJ3617 Best Cow Line
- poj3617 Best Cow Line
- POJ3617 Best Cow Line
- POJ3617-Best Cow Line
- poj3617 Best Cow Line
- poj3617 Best Cow Line
- POJ3617 Best Cow Line
- POJ3617 Best Cow Line
- poj3617 Best Cow Line 贪心
- POJ3617 Best Cow Line 贪心
- poj3617 Best Cow Line 贪心
- poj3617 Best Cow Line 贪心
- Java多线程的学习
- Java中finally语句块的深度解析(try catch finally的执行顺序)
- (最新)关于CocoaPods安装和使用
- SQL学习整理(二)检索数据
- javascript中canvas绘图的基本用法
- POJ3617 Best Cow Line
- 遮盖层加弹出框页面布局不影响
- mac下的pycharm可能遇到的问题
- 实体小三角-制作方法
- 中文乱码处理专题
- 动规+同余
- hihocoder #1033 交错和问题的思考
- 363.Trapping Rain Water-接雨水(中等题)
- Maven+Jetty/Tomcat进行web开发部署