Pat1067:Sort with Swap(0,*)

最新推荐文章于 2020-10-27 20:10:01 发布

转载最新推荐文章于 2020-10-27 20:10:01 发布 · 95 阅读

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原文链接：http://www.cnblogs.com/0kk470/p/7777192.html

本文介绍了一个特殊的排序问题——仅允许使用Swap(0,*)操作来对0到N-1的排列进行排序。通过贪心算法结合桶排序的方法，文章详细解释了如何找到最小的交换次数，并提供了实现这一算法的具体代码。

1067. Sort with Swap(0,*) (25)

时间限制

150 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Given any permutation of the numbers {0, 1, 2,..., N-1}, it is easy to sort them in increasing order. But what if Swap(0, *) is the ONLY operation that is allowed to use? For example, to sort {4, 0, 2, 1, 3} we may apply the swap operations in the following way:

Swap(0, 1) => {4, 1, 2, 0, 3}
Swap(0, 3) => {4, 1, 2, 3, 0}
Swap(0, 4) => {0, 1, 2, 3, 4}

Now you are asked to find the minimum number of swaps need to sort the given permutation of the first N nonnegative integers.

Input Specification:

Each input file contains one test case, which gives a positive N (<=10⁵) followed by a permutation sequence of {0, 1, ..., N-1}. All the numbers in a line are separated by a space.

Output Specification:

For each case, simply print in a line the minimum number of swaps need to sort the given permutation.

Sample Input:

10 3 5 7 2 6 4 9 0 8 1

Sample Output:

9

思路
贪心 + 桶排序。

1.用数组pos模拟N个桶记录输入的数的位置，数组的下标表示这个数。输入数字时根据输入次序i和数字大小num是否相等来决定需要和0交换的数字个数NumCount。
2.只要0没有在0位置，就将pos[0]与pos[pos[0]]交换.循环直到pos[0] == 0。
3.如果pos[0] == 0且还有数字不在正确的位置(NumCount > 0)，那么就将0和最近的一个不在正确位置的数字交换。注意索引交换的数字时用index记录下标（不然有几个测试用例会超时），这样下一次pos[0] == 0需要索引新的交换数字时不用再重头开始。
4.用一个变量swaptimes统计交换次数，然后输出就行。

代码

#include<iostream>
#include<vector>
using namespace std;
int main()
{
    int N;
    while(cin >> N)
    {
        int index = 1,NumCount = 0,swaptimes = 0;
        vector<int> pos(N);
        for(int i = 0;i < N;i++)
        {
            int num;
            cin >> num;
            pos[num] = i;
            if(pos[num] != num && num != 0)
                NumCount++;
        }
        while(NumCount > 0)
        {
            if(pos[0] == 0)
            {
                while(index < N)
                {
                    if(pos[index] != index)
                    {
                        swap(pos[0],pos[index]);
                        swaptimes++;
                        break;
                    }
                    index++;
                }
            }
            while(pos[0] != 0)
            {
                swap(pos[0],pos[pos[0]]);
                NumCount--;
                swaptimes++;
            }
        }
        cout << swaptimes << endl;
    }
}

转载于:https://www.cnblogs.com/0kk470/p/7777192.html