Best Cow Line
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
Source
题目类型：贪心
算法分析：如果头的字符比末尾的字符小，则输出头的字符，如果头的字符大于末尾的字符，则输出末尾的字符。如果相等，则需要找到第一个不相等的字符，然后输出这个字符是从头尾所走到的那个相同字符，注意最后输出的格式！！！
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#pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("aaa.txt", "r", stdin) #define CPPFF ifstream cin ("aaa.txt") #define LL long long #define ULL unsigned long long #define DB(ccc) cout << #ccc << " = " << ccc << endl const int INF = 0x7FFFFFFF; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 2e3 + 6; char aa[maxn]; int main() { // CFF; int n; while (scanf ("%d", &n) != EOF) { for (int i = 1; i <= n; i++) scanf (" %c", &aa[i]); int pa = 1, pb = n; int cnt = 0; while (pa <= pb) { bool is_left = false; for (int i = 0; pa + i <= pb; i++) { if (aa[pa+i] < aa[pb-i]) {is_left = true; break;} if (aa[pa+i] > aa[pb-i]) {is_left = false; break;} } if (cnt && cnt % 80 == 0) puts (""); if (is_left) printf ("%c", aa[pa++]); else printf ("%c", aa[pb--]); cnt++; } puts (""); } return 0; } |
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