poj3617 Best Cow Line 贪心

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.

题意：给你一个长度为N的字符串s，让你构造成一个字符串T，只有两种操作：1）从s的头部删除一个字符添加到T的尾部。

2）从s的尾部删除一个字符添加到T的尾部。要求T的字典序最小。

题解：这里用贪心的思想，头尾相比较，谁小放谁，如果一样的，那么比较后面的字符的大小。注意题目要求：字符满80个就要换行。。。

#include<iostream>#include<cstring>#include<cstdlib>#include<cstdio>using namespace std;int main(){    char c[2002];    int n,count;    while(cin>>n)    {        count=0;        int t=n-1;        for(int i=0;i<n;i++)        {            cin>>c[i];        }        int a=0;        int b=n-1;        while(a<=b)        {            bool flag=false;                for(int i=0;a+i<=b;i++)                {                    if(c[a+i]<c[b-i])                    {                        flag=true;                        break;                    }                    else if(c[a+i]>c[b-i])                    {                        flag=false;                        break;                    }                }            if(flag)                cout<<c[a++];            else                cout<<c[b--];                count++;            if(count==80)            {                cout<<endl;                count=0;            }        }        if(count)        cout<<endl;    }    return 0;}