A. Chess Placing

最新推荐文章于 2019-02-20 22:21:39 发布

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本文介绍了一种关于棋盘布局的优化算法问题，棋盘为1×n大小且n为偶数，初始放置若干棋子在黑白相间的格子上。目标是最小化移动次数使所有棋子位于同一颜色的格子上。通过排序和计算每一步移动的代价来求解。

A. Chess Placing
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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 .

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 integer numbers (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.

Examples
inputCopy
6
1 2 6
outputCopy
2
inputCopy
10
1 2 3 4 5
outputCopy
10
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 2 to 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 in 4 moves, then in 3 moves, in 2 moves and in 1 move.

#include<stdlib.h>
#include<stdio.h>
#include<algorithm>
using namespace std;
int a[101];
int main()
{
    int n;
    scanf("%d",&n);
    int i;
    for(i=1;i<=n/2;i++)
    {
        scanf("%d",&a[i]);
    }
    int sum1=0,sum2=0;
    sort(a+1,a+1+n/2);
    for(i=1;i<=n/2;i++)
    {
        sum1+=abs(a[i]-(2*i-1));
        sum2+=abs(a[i]-(2*i));
    }
    printf("%d\n",min(sum1,sum2));
    return 0;
}

题意：棋盘格是黑白相间的，往上放棋子，最终的棋子只能是在黑或者白上面，所以只可能是1，3，5，7，9…或者2，4，6，8，10…并且为什么要先排好序那，因为所有的棋子最后都会走到自己对应的2*i上的偶数或者2*i-1上的奇数上，我们这里减的是和i有关的，所以要对应你如果i=2的位置上放到是6,那6本来是移动到12上的，在2上就会移动到4上了，要的是对应，每个人找到自己的位置才能保证移动到对应的位置上