简单贪心算法——字典序最小问题（Best Cow Line POJ3617）

题目连接

http://poj.org/problem?id=3617

Best Cow Line POJ3617

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 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

题目限制

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 33926		Accepted: 8929

思路

题目为简单贪心思想，字典序问题可以联想贪心算法。从字符串的左端和右端分别取其字符进行字典序比较，选其小的字符加入到T 的末尾。若其相同则分别从两端向中间看下一位大小，直到找到可以区分大小的字符。如果全部相同，从左端还是右端取都可以。

代码

#include
using namespace std;

const int MAX_N = 2000;
//输入
int N ;
char S[MAX_N+1];
void solve()
{
    //剩余的字符串 s[a], s[a+1],s[a+2]...s[b]
    int  a = 0,b = N -1;
    int sum = 0;   //统计已经输出的字符个数
    while(a <= b)
    {
        bool left = false;  //left标志变量，标志哪一端的字典序小
        for(int i=0;a+i <= b;i++)
        {
            //如果相同则再次循环，直到找到字符大小不一样
            if(S[a+i] < S[b-i])
            {
                left = true;
                break;
            }

            else
                if(S[a+i]>S[b-i])
                {
                left = false;
                break;
                 }
        }

        //一旦从上个循环中跳出来，说明确定了字符的大小或者全部相同
        if(left)
        {       
            cout<>N;
    for(int i=0;i < N;i++)
    {
         cin>>S[i];
    }
     solve();
    return 0;
}

总结

贪心目标为两端最小的字符，字典序问题+贪心算法