Best Cow Line <挑战程序设计竞赛> 习题 poj 3617

P2870 [USACO07DEC]最佳牛线，黄金Best Cow Line, Gold
poj 3617 http://poj.org/problem?id=3617

题目描述
FJ is about to take his N (1 ≤ N ≤ 500,000) 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.

每次只能从两边取，要求取出来之后字典序最小

输入格式
* 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
输出格式
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.

样例

输入
6
A
C
D
B
C
B
输出
ABCBCD

简单的说就是从输入字符串队首队尾抽取字符串组成输出 要求输出字符串字典序最小

算法1
我们来贪心选取 队首队尾哪个字典序小 就选择哪个。
关键在队首队尾相同那么就看次一级的字母的字典序 但是要考虑诸如 aca vav aa 等边界情况。
需要处理好
吐槽下 POJ 的提交和没有错误提示。 洛谷倒是不错 就是数据量变大了 把我代码卡成TLE了

C++ 代码

#include <iostream>
#include <queue>
#include <string>

using namespace std;



char input[500010];
char output[500010];
int idx;

int  SelectCopy(int l, int r)
{
    if (l >= r) return 1;

    if (input[l] < input[r]) {
        return -1;
    }
    else if (input[l] > input[r]) {
        return  1;
    }
    else {
        int copyl = l + 1; int copyr = r - 1;
        return SelectCopy(copyl, copyr);
    }

    return 0;
}


void Select(int l, int r)
{
    if (l == r) { output[idx] = input[l]; idx++; return; }
    if (l > r) return;

    if (input[l] < input[r]) {
        output[idx] = input[l];
        idx++; l++;
    }
    else if (input[l] > input[r]) {
        output[idx] = input[r];
        idx++; r--;
    }
    else {
        //相等
        int copyl = l + 1; int copyr = r - 1;
        int selectidx = SelectCopy(copyl, copyr);
        if (-1 == selectidx) {
            output[idx] = input[l];
            idx++; l++;
        }
        else if (1 == selectidx) {
            output[idx] = input[r];
            idx++; r--;
        }
    }

    if (l <= r) {
        Select(l, r);
    }

}


int main()
{
    int N;
    cin >> N;
    for (int i = 0; i < N; i++) {
        char t;
        cin >> t;
        input[i] = t;
    }

    int l = 0; int r = N - 1;

    Select(l, r);
    for (int i = 0; i < idx; i++) {
        if (i % 80 == 0 && i>=80) cout << endl;
        cout << output[i];
    }

    return 0;
}

作者：defddr
链接：https://www.acwing.com/blog/content/178/
来源：AcWing
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