1067 Sort with Swap(0, i) (25分)

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 (≤) 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

题目分析：看浙大《数据结构》的时候见到过这题 重做并没有做成功 运行超时了 我想的是每次通过0与0的位置来归位 若0在这个过程中不小心被交换到了0的位置 那就得将0交换到还未被归为的那个元素上 思路是对的 做法导致时间复杂度过大
通过上面的分析 可以将0所在的看成一个环 只需要记录环中有多少元素 以及有多少环 当0所在的环计算完成后 就到另一个环去

过了3个测试点

#define _CRT_SECURE_NO_WARNINGS
#include <climits>
#include<iostream>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<algorithm>
#include<string>
#include<cmath>
using namespace std;
int Address[100001];
int Array[100001];
void swap(int i, int j)//交换2个地址上的值 i，j为地址
{
    Address[Array[i]] = j;
    Address[Array[j]] = i;
    int temp = Array[i];
    Array[i] = Array[j];
    Array[j] = temp;
}
int main()
{
    int N;
    int times = 0;
    cin >> N;
    for (int i = 0; i < N; i++)
    {
        cin >> Array[i];
        Address[Array[i]] = i;
    }
    int flag = 1;
    while (flag)
    {
        if (Address[0] == 0){
            for (int i = 1; i < N; i++)
                if (Address[i] != i) {
                    swap(Address[0], Address[i]);
                    flag = 1;
                    times++;
                    break;
                }
                else flag = 0;
        }
        else{
            swap(Address[0], Address[Address[0]]);
            times++;
        }            
    }
    cout << times;
}

全过

#define _CRT_SECURE_NO_WARNINGS  
#include<stdio.h>

int A[100000] = { 0 };
int Position[100000] = { 0 };
int IsRight[100000] = { 0 };
void Swap(int i, int j)
{
    int tmp = A[i];
    A[i] = A[j];
    A[j] = tmp;    
}
int SwapTimes=0;
int FindElements(int Pos)
{
    int num=1;
    while (Position[Pos]!=Pos)
    {
        num++;
        IsRight[Position[Pos]] = 1;
        Position[Pos] = Position[Position[Pos]];
    }
    return num;  //返回元素的个数
}
void Charge(int N)
{
    int num;
    //从零开始计算
    SwapTimes += FindElements(0)-1;            //交换次数比元素个数少一
    for (int i = 1; i < N; i++)
    {
        if (!IsRight[i])
            SwapTimes += FindElements(i)+1;        //虽然交换次数比元素个数少一 但是要利用0来进行交换 所以时 这个环的元素加一
    }                                            //而把0元素添加到 环中先进行一次交换 所以 最后结果为  元素+1-1+1
}
int main()
{
    int N;
    scanf("%d", &N);
    for (int i = 0; i < N; i++)
    {
        int num;
        scanf("%d", &num);
        A[i] = num;
        Position[num] = i;
        if (A[i] == i)
            IsRight[i] = 1;
    }
    Charge(N);
    printf("%d", SwapTimes);
    return 0;
}