uva 10570——Meeting with Aliens

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题意：输入1-n的排列，每次可以交换两个整数。用最少的次数把排列变成1-n的环状序列。

思路：枚举+贪心。依次枚举环上所有的点，正序一遍，倒序一遍，然后贪心求的所需的最小步数（贪心策略是1与1号交换，2与2号交换……）。

code：

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#include <sstream>#include <string>#include <vector>#include <list>#include <queue>#include <stack>#include <map>#include <set>#include <bitset>using namespace std;typedef long long ll;typedef unsigned long long ull;typedef long double ld;const int INF=0x3fffffff;const int inf=-INF;const int N=1000000;const int M=505;const int mod=1000000007;const double pi=acos(-1.0);#define cls(x,c) memset(x,c,sizeof(x))#define cpy(x,a) memcpy(x,a,sizeof(a))#define fr(i,s,n) for (int i=s;i<=n;i++)#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define lrt  rt<<1#define rrt  rt<<1|1#define middle int m=(r+l)>>1#define lowbit(x) (x&-x)#define pii pair<int,int>#define mk make_pair#define IN freopen("in.txt","r",stdin);#define OUT freopen("out.txt","w",stdout);int n,vis[M],v[M*2];int sol(int *p){    cls(vis,0);    int ct=0,k;    fr (i,1,n)    {        if (vis[i]) continue;        ct++;k=i;        while (!vis[k]) vis[k]=1,k=p[k];    }    return n-ct;}int  main(){    while (~scanf("%d",&n)&&n)    {        fr (i,1,n) scanf("%d",&v[i]);        int ans=INF;        fr(i,1,2)        {            fr (j,1,n) v[j+n]=v[j];            fr(j,1,n) ans=min(ans,sol(v+j));            reverse(v+1,v+1+n);        }        printf("%d\n",ans);    }}

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