POJ3617 Best Cow Line（贪心:字典序最小+正确的输入/输出姿势）

存入草稿箱的日期为1.29，咕咕咕
打jjc的时候服务器炸了，gww退钱！
Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 36220

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

6
A
C
D
B
C
B

Sample Output

ABCBCD

Solution

道理很简单，就是将开头和末尾比较，取最小的加入T中。
由于会出现首位相同的情况，所以我们需要比较下一位。
然而下一位也可能相同。
这个时候就需要将S倒序比较，谁字典序小就取谁的首位。
虽然已经很明白了，但是自己写是不可能会写的，只能照着小白书上的背下来这样勉强生活。
学习之后：
坑有、多，见下
·单个输入
可以使用代码中的方式（上网搜的），也可以使用scanf(" %d“，&a);空格代表换行符、tab等
·输出时a[head++]=={a[head];++head}；并且每80个字符换行

AC代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
int main()
{
 int n;
 while(~scanf("%d",&n))
 {
  string a="";
  char x;
  for(int i=0;i<n;++i)
  { 
   cin>>x;
   a=a+x;
  }
  int tail=0,head=n-1;
  int sum=0;
  while(tail<=head)
  {
   bool b=0;
   for(int i=0;tail+i<=head;++i)
   {
    if(a[tail+i]<a[head-i])
    {
     b=1;
     break;
    }
    else if(a[tail+i]>a[head-i])
    {
     b=0;
     break;
    }
   }
   if(b)
   {
    printf("%c",a[tail]);
    ++tail;
   }
   else
   {
    printf("%c",a[head]);
    --head;
   }
   ++sum;
   if(sum%80==0) printf("\n");
  }
 }
 return 0;
}