1067. 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

这道题目正着记录还不行，一定要换一种思路，数组记录的是每个数字的位置，而不是每个位置放的是什么数字

#include <cstdio>
#include <vector>
using namespace std;
int main() {
int count, index = 0, diff = 0, ans = 0;
int data[100000];
scanf("%d", &count);
for(int i = 0; i < count; i++){
int num;
scanf("%d", &num);
data[num] = i;
if(num != i)
diff++;
}

while(diff > 0){
if(data[0] != 0){
int temp = data[0];
data[0] = data[data[0]];
data[temp] = temp;
ans++;
diff--;
if(data[0] == 0)
diff--;
}
else{
for(int i = index; i < count; i++){
if (i != data[i]){
index = i;
data[0] = data[i];
data[i] = 0;
ans++;
diff++;
break;
}
}
}
}
printf("%d\n", ans);
return 0;
}