poj 3617 Best Cow Line

Best Cow Line

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 19186		Accepted: 5319

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

题目大意：给定长度为N的字符串S，要构造一个长度为N的字符串T。起初，T是一个空串，随后反复进行下列任意操作。

从S的头部删除一个字符，加到T的尾部

从S的尾部删除一个字符，加到T的尾部

目标是构造字典序尽可能小的字符串。

 

题目分析；从字典序的性质上看，无论T的末尾有多大，只要前面部分较小就可以。所以我们可以试一下如下贪心算法：

不断取S和T的末尾中较小的一个字符放到T的末尾

 

这个算法已经接近正确了，只是针对S的开头和末尾字符相同的情形还没有定义。在这种情况下，因为我们希望能够尽早使用更小的字符，所以就要比较一下下一个字符的大小。下一个字符也有可能相同，因此就有如下算法：

按照字典序比较字符串S和S反转后的字符串S'。

如果S较小，就从S的开头取出一个文字，放到T的末尾。

如果S'较小，就从S的末尾取出一个文字，放到T的末尾。

（若果相同则去哪一个都可以）

 

#include <cstdio>
int n;
char s[2002], temp[2];

void solve(){
    int a = 0, b = n - 1, cnt = 0, i;
    bool left = false;
    while(a <= b){
        cnt++;
        for(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--]);
        if(cnt % 80 == 0)
            printf("\n");
    }

}

int main(){
    scanf("%d", &n);
    for(int i = 0; i < n; i++){
        scanf("%s", temp);
        //若scanf("%c", &s[i])，则是错的，因为回车也是一个字符 
        s[i] = temp[0];
    }
    solve();
    return 0;
}

#include <iostream>
#include <cstdio>
using namespace std;
int n;
char s[2002];

void solve(){
    int a = 0, b = n - 1, cnt = 0;
    bool left = false;
    while(a <= b){
        for(int i = 0; a+i <= b-i; 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++;
                //如果一行满80个字符了，则换行
        if(cnt % 80 == 0)
            printf("\n");
    }
        //如果cnt % 80 == 0则说明在之前换过行了，不用换行，否则要换行
    if(cnt % 80)
        printf("\n");
}

int main(){
    while(cin >> n){
        for(int i = 0; i < n; i++){
            cin >> s[i];
        }
        solve();
    }
    return 0;
}