pat1067. Sort with Swap(0,*) (25)
1067. Sort with Swap(0,*) (25)
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 1Sample Output:
9
思路：将序列转换为图，形成键-值关系
例如样例：3 5 7 2 6 4 9 0 8 1
对应的值：0 1 2 3 4 5 6 7 8 9
对应的键：3 5 7 2 6 4 9 0 8 1
发现3类环：
a.元素个数>1，包含0的环：0-7-2-3-0
b.元素个数>1，不包含0的环：1-9-6-4-5-1
c.元素个数==1的环：8
其实一张图无非也就最多存在上述3类环。
对于a：交换次数=环的元素个数-1
对于b：交换次数=环的元素个数+1
对于c：交换次数=0
所以，只要统计元素个数>1的环的情况即可。然后根据这些元素个数>1的环是否包含元素0来计算总的交换次数。
代码如下：
1 #include<cstdio> 2 #include<stack> 3 #include<cstring> 4 #include<iostream> 5 #include<stack> 6 #include<set> 7 #include<map> 8 using namespace std; 9 map<int,int> ha; 10 bool vis[100005]; 11 void DFS(int cur,int s){ 12 vis[cur]=true; 13 if(ha[cur]==s){ 14 return; 15 } 16 DFS(ha[cur],s); 17 } 18 //统计点的数量>1的环的个数，有0在的换交换次数是个数-1；没有0在的环，交换次数是个数+1 19 int main(){ 20 //freopen("D:\\INPUT.txt","r",stdin); 21 int count,n; 22 scanf("%d",&n); 23 count=n; 24 int i,num; 25 for(i=0;i<n;i++){ 26 scanf("%d",&num); 27 if(num==i){//去掉元素个数为0的环，剩下的元素个数 28 vis[num]=true; 29 count--; 30 } 31 ha[num]=i; 32 } 33 //剩下的环除了包含元素0的环，其他环的交换次数=环自身元素个数+1 34 for(i=0;i<n;i++){ 35 if(!vis[i]){ 36 DFS(i,i); 37 count++; 38 } 39 } 40 if(ha[0]!=0){//如果0不单独成环，说明有元素个数>1的环包含0，则这个环在34行时多加了2，这里减去 41 count-=2; 42 } 43 printf("%d\n",count); 44 return 0; 45 }
