POJ 3617 – Best Cow Line [贪心+枚举]

Link

传送门

Problem

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.

Mean

N个字符,每次取首或尾最终形成一个新字符串,使得新串字典序最小

Analysis

用两个指针,暴力判断每次取时首或尾哪个是最优的。判断时两个指针同时往中间走,直到走到不同为止,哪边小取哪边。如果相同,那么随便取一个。

因为数据量不大,这样直接暴力O(N^2)完全可以。

Code

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int maxn = 2048;
char v[maxn];
string s;

bool judge(int L, int R) {
    for(int i = 0; L + i < R - i; ++i) {
        if(v[L+i] < v[R-i]) return false;
        if(v[L+i] > v[R-i]) return true;
    }
    return false;
}

int main()
{
    ios::sync_with_stdio(false);
    int N; cin >> N;
    for(int i = 1; i <= N; ++i)
        cin >> v[i];
    int L = 1, R = N;
    while (s.size() < N) {
        if(!judge(L, R)) s.push_back(v[L++]);
        else s.push_back(v[R--]);
    }
    for(int i = 0; i < N; ++i) {
        cout << s[i];
        if((i + 1) % 80 == 0)
            cout << endl;
    }
    if(N % 80) cout << endl;
    return 0;
}

欢迎留言