Best Cow Line(POJ 3617) 贪心

来自《挑战程序设计竞赛》

1.题目原文

http://poj.org/problem?id=3617

Best Cow Line

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 19765 Accepted: 5457

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

给定一个长度为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，随后反复进行如下操作:

1.从S的头部删除一个字符，加到T的尾部。

2.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T。

2.解题思路

字典序比较类的问题经常能用到贪心算法。

我们可以试一下，如下贪心算法:不断取S的头部和尾部中较小的字符放到T的尾部。

这个算法几乎是正确的，但是对于S的开头和末尾字符相同的情况没有定义。在这种情况下，我们想尽可能早点使用较小的字符，就需要比较下一个字符，下一个字符也可能相同。因此，就有如下算法:

按照字典序比较S和S反转后的字符串S'。如果S较小，就把S开头的字符加到T末尾。如果S’较小，就把S末尾的字符加到T末尾。如果相同，加入哪一个都可以。

3.AC代码

#include<algorithm>#include<cctype>#include<cmath>#include<cstdio>#include<cstdlib>#include<cstring>#include<iomanip>#include<iostream>#include<map>#include<queue>#include<string>#include<set>#include<vector>#include<cmath>#include<bitset>#include<stack>#include<sstream>#include<deque>#include<utility>using namespace std;const int maxn=2005;int n;char s[maxn];void solve(){    int a=0,b=n-1;    int sum=0;    while(a<=b){        //将从左起和右起的字符串进行比较        bool left=false;        for(int 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;            }        }        sum++;        if(left) cout<<s[a++];        else cout<<s[b--];        if(sum%80==0){            cout<<endl;        }    }}int main(){    while(scanf("%d",&n)!=EOF){        for(int i=0;i<n;i++){            cin>>s[i];        }        solve();    }    return 0;}