解题报告poj1674交换排序.doc

Sorting by Swapping Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 5760 Accepted: 3033 Description Given a permutation of numbers from 1 to n, we can always get the sequence 1, 2, 3, ..., n by swapping pairs of numbers. For example, if the initial sequence is 2, 3, 5, 4, 1, we can sort them in the following way: 2 3 5 4 1 1 3 5 4 2 1 3 2 4 5 1 2 3 4 5 Here three swaps have been used. The problem is, given a specific permutation, how many swaps we needs to take at least. Input The first line contains a single integer t (1 = t = 20) that indicates the number of test cases. Then follow the t cases. Each case contains two lines. The first line contains the integer n (1 = n = 10000), and the second line gives the initial permutation. Output For each test case, the output will be only one integer, which is the least number of swaps needed to get the sequence 1, 2, 3, ..., n from the initial permutation. Sample Input 2 3 1 2 3 5 2 3 5 4 1 Sample Output 0 3 参考了别人的算法。。 {3 ,7 ,1 ,6 ,2 ,4 ,8 ,5 }再和标准序列 {1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 }比较 找出其中所有的环 这里的环就是指它们互相交换之后能成为标准序列的最小集合 比如 这里{1,3}是一个环, {7,2,5,8}是一个环。 具体找法很简单 首先确定一个不在已找出环中的数字 例如第一次从3开始,3对应标准序列的1 再找1对应的标准序列3 3回到了开始的数字 那么这个环就是{1,3} 第二次从7开始,7-2 2-5 5-8 8-7 所以第二个环是{7,2,5,8} 第三个环是{6,4} 这样所有的数字都在环中了 交换的次数=(数字总数-环数)=8-3=5#include stdio.h int main() { int m,n,i,j,k,a[10010]={0},b[10010]={0},sum; scanf(%d,m); while(m--) { scanf(%d,n); sum=0; for(i=0;in;i++)scanf(%d,a[i]); for(i=0;in;i++)b[i]=i+1; for(i=0;in;i++) { j=i; if(b[i]!=0) { while(b[i]!=a[j]) { j=a[j]-1; b[j]=0; } sum++; } } printf(%d\n,n-sum); } return 0; }