poj3617 Best Cow Line

標籤：poj

轉載請註明出處：http://blog.csdn.net/u012860063

題目連結：http://poj.org/problem?id=3617

【題意】

一個長度為N（N<=1000）的字串，每次可以從隊尾或隊首拿出一個字元加入到新字串隊尾，求字典序最小的新字串

【輸入】

第一行一個N

接下來N行每行一個大寫字母

【輸出】

字典序最小的新字串

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

6ACDBCB

Sample Output

ABCBCD

Source

USACO 2007 November Silver

代碼如下：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define PI acos(-1.0)#define INF 0x3fffffffint main(){int n;char c,a[2047];int i,k,j;while(~scanf("%d",&n)){k = 0 ;memset(a,0,sizeof(a));for(i = 0; i < n; i++){getchar();scanf("%c",&a[i]);}int t1=0,t2=n-1;//t1是首部t2是尾部for(i = 0; ; i++){if(a[t1] < a[t2]){printf("%c",a[t1]);t1++,k++;}else if(a[t1] > a[t2]){printf("%c",a[t2]);t2--,k++;}else if(a[t1] == a[t2])//如果首尾的字母是相同的{//就接著向中間找，直至找到兩個不同的int c1 ,c2;c1= t1,c2=t2;while(a[c1] == a[c2]){c1++,c2--;}if(a[c1] < a[c2]){printf("%c",a[t1]);t1++,k++;}else{printf("%c",a[t2]);t2--,k++;}}if(k%80 == 0)printf("\n");if(k == n)break;}printf("\n");}return 0;}