POJ--2386 Lake Counting(DFS)

记录
0:33 2023-1-24

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

reference:《挑战程序设计竞赛(第2版)》2.2.3 p43

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

一个贪心算法,书上描述:

需要注意的就是POJ上，这个题要求一行80个字符，多了要换行，不然会presentation error，也就是格式错误。(有种感觉其它题输出也可能要求这样，80个字符，大概是老式terminal？console？啥的一行的字符吧，这里逻辑有点乱。。)

#include<cstdio>
#define MAX_N 10000

int N;
char S[MAX_N + 1];

void solve() {
    // 剩余字符串为S[a], S[a+1], ...., S[b]
    int a = 0, b = N - 1;
    // POJ specification
    // if one line excedd 80 characters, must change new line
    int num = 0;

    while (a <= b) {
        // 左起与右起字符串进行比较
        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;
            }
        }

        if (left) putchar(S[a++]);
        else putchar(S[b--]);

        num++;
        if(num == 80) {
            num = 0;
            putchar('\n');
        }
    }
}

int main() {
    char c;
    scanf("%d\n", &N);
    for(int i = 0; i < N; i++) {
        scanf("%c", &c);
        S[i] = c;
        getchar();
    }
    solve();
    // char c;
    // while (~scanf("%d\n", &N)) {
    //     for(int i = 0; i < N; i++) {
    //         scanf("%c", &c);
    //         S[i] = c;
    //         getchar();
    //     }
    //     solve();
    // }
}