1067. Sort with Swap(0,*) (25)

时间限制100 ms内存限制32000 kB代码长度限制16000 B判题程序Standard作者CHEN, YueGiven 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 followingway: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

分析：

（1）用正常思路做的话，还是会超时间，需要剪枝，或者用类似并查集的方法。

（2）剪支的话，内层循环，从K开始就好，K是每次记录的上一次循环到的地方。因为之前检查如果不成立的话，下一次检查还是不成立。

（3）visit[]可以一个表达式写出来，就不用单独列了。

#include<cstdio>#include<vector>using namespace std;int ip[100010],pos[100010],n;void sort (int a,int b){int tmp=pos[a];pos[a]=pos[b];pos[b]=tmp;}int main(){scanf("%d",&n);int i,j;for (i=0;i<n;i++){scanf("%d",&ip[i]);pos[ip[i]]=i;}int cnt=0;int k=1;for (i=0;i<n;i++){if (pos[pos[0]]==0){for (j=k;j<n;j++){if (pos[j]!=j&&j!=pos[0]){k=j;sort(0,j);cnt++;break;}}}{if(pos[0]==0)break;sort(0,pos[0]);cnt++;}}printf("%d",cnt);///return 0;}