POJ-3617 Best Cow Line （贪心）

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描述

传送门：POJ-3617 Best Cow Line

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.

输入描述

  • 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

输出描述

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.

示例

输入

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A
C
D
B
C
B

输出

1
ABCBCD

题解

题目大意

给定一个长度为N的字符串S，T是一个空串，每次从头部或尾部删除一个字符加到T的尾部，构造字典序最小的字符串T。

思路

不断取开头或末尾中较小的一个字符到T中，注意两边相等的情况，还有坑点是输出格式是80行换行。

代码

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#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <string>
#include <cstring>
#include <queue>
const int MAXN = 2005;
using namespace std;
char s[MAXN];

int main(){
    int n;
    while(cin >> n){
        for(int i = 0; i < n; i++){
            cin >> s[i];
        }
        int a = 0, b = n-1, cnt = 0;
        while(a <= b){
            bool left = false;
            for(int i = 0; a+i <= b; i++){
                if(s[a+i] < s[b-i]){
                    left = true;
                    break;
                }
                else if(s[a+i] > s[b-i]){
                    left = false;
                    break;
                }
            }
            if(left) putchar(s[a++]);
            else putchar(s[b--]);
            cnt++;
            if(cnt == 80){
                cnt=0;   
                cout << endl;
            }
        }
        if(cnt)
            cout << endl;
    }
}