pat1067. 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

---

提交代码

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：

 #include<cstdio>
 #include<stack>
 #include<cstring>
 #include<iostream>
 #include<stack>
 #include<set>
 #include<map>
 using namespace std;
 map<int,int> ha;
 bool vis[];
 void DFS(int cur,int s){
     vis[cur]=true;
     if(ha[cur]==s){
         return;
     }
     DFS(ha[cur],s);
 }
 //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
 int main(){
     //freopen("D:\\INPUT.txt","r",stdin);
     int count,n;
     scanf("%d",&n);
     count=n;
     int i,num;
     for(i=;i<n;i++){
         scanf("%d",&num);
         if(num==i){//去掉元素个数为0的环，剩下的元素个数
             vis[num]=true;
             count--;
         }
         ha[num]=i;
     }
     //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
     for(i=;i<n;i++){
         if(!vis[i]){
             DFS(i,i);
             count++;
         }
     }
     if(ha[]!=){//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
         count-=;
     }
     printf("%d\n",count);
     return ;
 }