Educational Codeforces Round 44 (Rated for Div. 2) A. Chess Placing

本文链接：https://blog.youkuaiyun.com/ACMer_QJ/article/details/80484990

探讨了在1*n大小的特殊棋盘上，如何通过最少步骤将棋子全部移至相同颜色格子的问题。文章提供了算法思路及C++实现代码。

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You are given a chessboard of size 1 × n. It is guaranteed that n is even. The chessboard is painted like this: "BWBW...BW".

Some cells of the board are occupied by the chess pieces. Each cell contains no more than one chess piece. It is known that the total number of pieces equals to $\text{[math]}$ .

In one step you can move one of the pieces one cell to the left or to the right. You cannot move pieces beyond the borders of the board. You also cannot move pieces to the cells that are already occupied.

Your task is to place all the pieces in the cells of the same color using the minimum number of moves (all the pieces must occupy only the black cells or only the white cells after all the moves are made).

Input

The first line of the input contains one integer n (2 ≤ n ≤ 100, n is even) — the size of the chessboard.

The second line of the input contains $\text{[math]}$ integer numbers $\text{[math]}$ (1 ≤ pi ≤ n) — initial positions of the pieces. It is guaranteed that all the positions are distinct.

Output

Print one integer — the minimum number of moves you have to make to place all the pieces in the cells of the same color.

Note

In the first example the only possible strategy is to move the piece at the position 6 to the position 5 and move the piece at the position 2to the position 3. Notice that if you decide to place the pieces in the white cells the minimum number of moves will be 3.

In the second example the possible strategy is to move $\text{[math]}$ in 4 moves, then $\text{[math]}$ in 3 moves, $\text{[math]}$ in 2 moves and $\text{[math]}$ in 1 move.

题意：给你一个1*n（n偶数）的棋盘，给你n/2个棋子下标，让你把所有棋子全放到奇数位置或全放到偶数位置，棋子平移到相邻位置花费为一，求最小花费。

思路：只可能放到奇数位或偶数位，且一一对应，先排序，分别算两种情况对应位置与排好序的初始下标相减的绝对值之和即为两种情况的最优解，再取min（奇数位花费，偶数位花费）即为答案。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;
#define ll long long
const int maxn = 105;

int main()
{
        int n, a[maxn], odd = 0, even = 0;
        scanf("%d", &n);
        for (int i = 1; i <= n/2; i ++) {
                scanf("%d", &a[i]);
        }
        sort(a+1, a+n/2+1);
        for (int i = 1, temp = 1; i <= n/2; i ++, temp += 2) {
                odd += abs(a[i] - temp);
                even += abs(a[i] - temp - 1);
        }
        printf("%d\n", min(odd, even));
}