POJ 3617 Best Cow Line 貪心

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Best Cow Line

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9116		Accepted: 2762

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
題解

題目要求是，有兩種操作，從頭取出一個字元，或者從尾取出一個字元，加到一個新的字串末端，然後保證這個字串是所有可能產生的字串中，字典序排列最小的那個。依舊是貪心的想法，誰小我取誰即可。特殊判斷兩者相同的。一旦兩者相同，那麼就接著判斷下一種，直到找到更小的為止。

另外需要注意兩點，題目的輸入很變態，要一個字母一行的輸入，所以用cin或者用scanf(" %c", &ch)。cin不多說了，除了慢點其他沒啥缺點。scanf中為何要加上一個空格呢？因為在格式串中，空格的意思是匹配輸入中的所有換行、TAB、空格，所以加上一個空格，就可以屏蔽掉在輸入中的所有的不愉快的因素了。

還有一個就是輸出，切記，每80個字元輸出一行。所以不要犯多輸出一行空行的錯誤。

程式碼範例

/**============================================================================#       COPYRIGHT NOTICE#       Copyright (c) 2014 All rights reserved#       ----Stay Hungry Stay Foolish----##       @author       :Shen#       @name         :POJ 3617#       @file         :G:\My Source Code\【ACM】訓練\0624 - 基礎\poj3617.cpp#       @date         :2014/06/24 13:46#       @algorithm    :Greedy============================================================================**///#pragma GCC optimize ("O2")//#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cmath>#include <cstdio>#include <string>#include <cstring>#include <iomanip>#include <iostream>#include <algorithm>using namespace std;template<class T>inline bool updateMin(T& a, T b){ return a > b ? a = b, 1: 0; }template<class T>inline bool updateMax(T& a, T b){ return a < b ? a = b, 1: 0; }typedef long long int64;int n;char t[2005], s[2005];void solve(){    for (int i = 0; i < n; i++)        scanf(" %c", &t[i]);    int i = 0, j = n - 1;    for (int k = 0; k < n; k++)    {        if (t[i] > t[j]) s[k] = t[j--];        else if (t[i] < t[j]) s[k] = t[i++];        else        {            int it = i, jt = j;            bool flag = 0;            while (t[it] == t[jt] && it <= jt) it++, jt--;            flag = (t[it] < t[jt]);            if (flag) s[k] = t[i++];            else s[k] = t[j--];        }    }    for (int i = 0; i < n; i++)    {        printf("%c", s[i]);        if (i % 80 == 79) printf("\n");    }    if (n % 80) printf("\n");}int main(){    while (~scanf("%d", &n)) solve();    return 0;}