Best Cow Line POJ - 3617（贪心)

本是古典何须时尚 2022-03-10 09:38 166阅读 0赞

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 *i*th 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

解题思路:

贪心策略,将字符串左位置元素与字符串右位置元素比较,将较小的那个串到新的字符串上去,

注意每输出80个字符要换行。。。

这题的难点是左右相等的情况,这时我们要进一步比较,我这一题在这个点超时了

我们在while里再套用一个for循环进行比较,循环直到左右两字符不再相等为止,这样就不会超时了

#include<cstdio>
    #include<algorithm>
    #include<cstring>
    #include<cmath>
    using namespace std;
    const int MAXN = 2010;
    int N;
    char tstr[MAXN];
    int main() {
    	while (scanf("%d", &N) != EOF) {
    		getchar();
    		for (int i = 0; i < N; ++i) {
    			scanf("%c", &tstr[i]);
    			getchar();
    		}
    		int pointL = 0;
    		int pointR = N - 1;
    		int index = 0;
    		int tcount = 0;
    		while (pointL <= pointR) {
    			if (pointL == pointR) {
    				printf("%c", tstr[pointL]);
    				tcount++;
    				break;
    			}
    			bool isLeft = false;
    			for (int i = 0; pointL + i <= pointR; ++i) {
    				if (tstr[pointL + i] < tstr[pointR - i]) {
    					isLeft = true;
    					tcount++;
    					break;
    				}
    				else if (tstr[pointL + i] > tstr[pointR - i]) {
    					isLeft = false;
    					tcount++;
    					break;
    				}
    			}
    
    			if (isLeft) putchar(tstr[pointL++]);
    			else putchar(tstr[pointR--]);
    			if (tcount % 80 == 0) printf("\n");  //每输出80个字符要换行
    		}
    		putchar('\n');
    	}
    	system("PAUSE");
    	return 0;
    }