POJ 3617 Best Cow Line 贪心

Best Cow Line

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 14492 Accepted: 4124

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

[Submit]   [Go Back]   [Status]   [Discuss]

给定一个串str 每次进行两个操作

一：从str的头部删除一个字母加到t串

二：从str的尾部删除一个字母加到t串

要求构造字典序最小的串，如果str[a]<str[b]删str[a]；如果str[a]>str[b]删str[b]相等继续往里面搜索

ACcdoe;

#pragma warning(disable:4786)//使命名长度不受限制#pragma comment(linker, "/STACK:102400000,102400000")//手工开栈#include <map>#include <set>#include <queue>#include <cmath>#include <stack>#include <cctype>#include <cstdio>#include <cstring>#include <stdlib.h>#include <iostream>#include <algorithm>#define rd(x) scanf("%d",&x)#define rd2(x,y) scanf("%d%d",&x,&y)#define rds(x) scanf("%s",x)#define rdc(x) scanf("%c",&x)#define ll long long int#define maxn 100005#define mod 1000000007#define INF 0x3f3f3f3f //int 最大值#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)#define MT(x,i) memset(x,i,sizeof(x))#define PI  acos(-1.0)#define E  exp(1)using namespace std;char str[maxn];int main(){    int n,a,b;    while(rd(n)!=EOF){        FOR(i,1,n)cin>>str[i];        bool flag;        b=n,a=1;        int cnt=0;        while(a<=b){            for(int i=0;a+i<b;++i)            if(str[a+i]<str[b-i]){                flag=true;break;            }else if(str[a+i]>str[b-i]){                flag=false;break;            }            if(flag)putchar(str[a++]);            else putchar(str[b--]);            cnt++;            if(cnt%80==0)putchar('\n');        }        printf("\n");    }    return 0;}/*6ACDBCB*/