A1067. Sort with Swap(0,*) (25)

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

索引数字反向存储，每次枚举第一个位置是否为0，为0在枚举未归位后是否有数字，没有则说明已排好

#include<cstdio>#include<algorithm>using namespace std;const int maxn = 1e5 + 10;int main(){int a[maxn];int n, temp, k = 1, ans = 0;//k保存未归位数的最小值  ans保存交换次数 scanf("%d", &n);int count = n - 1;for(int i = 0; i < n; ++i){scanf("%d", &temp);a[temp] = i;} while(1){while(a[0] != 0){swap(a[0], a[a[0]]);//交换0与a[0] ans++;count--;}while(k < n){if(a[k] != k){swap(a[0], a[k]);//交换0与k的位置ans++;break;}++k;}if(k == n)      //k到达最后一个数的下一位置，说明都已归位 break;}printf("%d\n", ans);return 0;}/*    //殊途同归1using namespace std;const int maxn = 1e5 + 10;int a[maxn], ans, n, k;int main(){ans = 0, k = 1;int temp;//temp接收索引, k 代表未归位的最小数 scanf("%d", &n);int count = n - 1;for(int i = 0; i < n; ++i){scanf("%d", &temp);a[temp] = i;if(temp != 0 && temp == i)count--;}while(count > 0){while(a[0] != 0){temp = a[0];swap(a[0], a[temp]);ans++; }for(int i = k; i < n; ++i){if(a[i] != i){swap(a[0], a[i]);k = i;ans++;break;}if(i == n-1){printf("%d\n", ans);return 0;}}}return 0;}*/
/*           //殊途同归2int main(){ans = 0, k = 1;// k 代表未归位的最小数 scanf("%d", &n);int temp, count = n - 1;//temp接收索引for(int i = 0; i < n; ++i){scanf("%d", &temp);a[temp] = i;if(temp != 0 && temp == i){count--;}}while(count > 0){if(a[0] == 0){while(k < n){if(a[k] != k){swap(a[0], a[k]);ans++;break;}++k;}}while(a[0] != 0){swap(a[0], a[a[0]]);ans++;count--;}}printf("%d\n", ans);return 0;}*/