POJ —— 3617 Best Cow Line

Best Cow Line
Time Limit: 1000MS
 
Memory Limit: 65536K
Total Submissions: 14395
 
Accepted: 4096
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

6
A
C
D
B
C
B

Sample Output
ABCBCD
貪欲な思想:
配列の先頭と末尾の小さな文字から出力され続けます.
このように議論が冒頭と末尾が同じ場合を忘れてしまうので,さらに最適化するアルゴリズムは以下のようになる.
a配列とa配列の反転後の文字列a′を辞書順に比較する.
aが小さい場合は、aの先頭から文字を取り出し、出力します.
a'が小さい場合は、aの末尾から文字出力を取り出します.
#include #include #include #include #include #define M(i,n,m) for(int i = n;i < m;i++) #define N(n,m) memset(n,m,sizeof(n)); const int MAX = 2002; using namespace std; char a[MAX]; int n; void solve(){int count=0;//出力時の空白行について議論int x=0,y=n-1;while(x<=y)//x=yは、この時点でx個またはy個の数が出力された可能性があることをx+i<=yで示し、この領域の数については議論する必要はない{//同じなら、異なる出力の小さいif(a[x+i] a[y - i])             {                 left = false;                 break;             }         }         if(left)         {             printf("%c",a[x++]);             count++;             if(count % 80 == 0)                 printf("");         }         else         {             printf("%c",a[y --]);             count++;             if(count % 80 == 0)                 printf("");         }     }     printf("");     return ; } int main() {     while(cin>>n)     {         M(i,0,n)         cin>>a[i];         solve();     }     return 0; }