pat1067. Sort with Swap (0,*) (25)

1067. Sort with Swap (0,*) (25) time limit MS Memory limit 65536 KB code length limit 16000 B award Program StandardAuthor Chen, Yue

Given any permutation of the numbers {0, 1, 2,..., N-1}, it's easy-to-sort them in increasing order. What if swaps (0, *) are the only operation and that's allowed to use? For example, to sort {4, 0, 2, 1, 3} We'll apply the swap operations in the following:

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 is asked to find the minimum number of swaps need to sort the given permutation of the first N nonnegative intege Rs.

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 is 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

Submit Code

Idea: Converting a sequence to a graph to form a key-value relationship

Examples: 3 5 7 2 6 4 9 0 8 1

Corresponding values: 0 1 2 3 4 5 6 7 8 9

corresponding keys: 3 5 7 2 6 4 9 0 8 1

3 Types of rings found:

A. Number of elements >1, containing 0 rings: 0-7-2-3-0

B. Number of elements >1, not including 0 rings: 1-9-6-4-5-1

C. Number of elements ==1 ring: 8

In fact, a picture is nothing more than the existence of the above 3 types of rings.

For a: number of switches = number of elements of the ring-1

For B: Number of switches = number of elements in the loop +1

For C: Interchange number =0

So, as long as the number of elements >1 the loop of the case can be. The total number of interchanges is then calculated based on whether the rings of these elements >1 contain element zeros.

The code is as follows:

1#include <cstdio>2#include <stack>3#include <cstring>4#include <iostream>5#include <stack>6#include <Set>7#include <map>8 using namespacestd;9map<int,int>ha;Ten BOOLvis[100005]; One voidDFS (intCurints) { Avis[cur]=true; -     if(ha[cur]==R) { -         return; the     } - DFS (ha[cur],s); - } - //The number of statistical points >1 rings, there are 0 in the number of exchanges is 1; no 0 in the ring, the number of exchanges is +1 + intMain () { -     //freopen ("D:\\input.txt", "R", stdin); +     intCount,n; Ascanf"%d",&n); atCount=N; -     intI,num; -      for(i=0; i<n;i++){ -scanf"%d",&num); -         if(num==i) {//Remove the number of elements 0, the number of elements remaining -vis[num]=true; incount--; -         } toha[num]=i; +     } -     //The rest of the ring, in addition to the ring containing element 0, the number of exchanges of the other rings = the number of elements of the ring +1 the      for(i=0; i<n;i++){ *         if(!Vis[i]) { $ DFS (i,i);Panax Notoginsengcount++; -         } the     } +     if(ha[0]!=0){//If 0 is not a separate ring, indicating that there is an element number >1 the ring contains 0, then this ring in 34 lines add 2, here minus Acount-=2; the     } +printf"%d\n", count); -     return 0; $}

pat1067. Sort with Swap (0,*)