【习题 8-13 UVA - 10570】Meeting with Aliens
2018-01-11 15:39
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【链接】 我是链接,点我呀:)
【题意】
在这里输入题意
【题解】
枚举1的位置在i
往右摆成一排。
a[i+1]..a
..a[1]..a[i-1]变为有序的
->寻找循环节,每个循环节的长度-1是必要的步骤数
->获取必要的步骤数,取最小值。
->O(n^2)
往左排成一排
->同样的方法处理就好
【代码】
#include <bits/stdc++.h> using namespace std; const int N = 5e2; int a[N+10],n,cnt; bool flag[N+10]; vector <int> v; int get_step(){ memset(flag,0,sizeof flag); int now = 0; for (int i = 1;i <= n;i++) if (!flag[i]){ int cc = 0,kk = i; while (!flag[kk]){ flag[kk] = true; cc++; kk = v[kk]; } now+=cc-1; } return now; } int main(){ #ifdef LOCAL_DEFINE freopen("rush_in.txt", "r", stdin); #endif ios::sync_with_stdio(0),cin.tie(0); while (cin >>n && n){ for (int i = 1;i <= n;i++) cin >> a[i]; int j = 1; for (int i = 2;i <= n;i++) if (a[i]==1) j = i; int ans = -1; v.resize(n+1); for (int i = 1;i <= n;i++){ cnt = 0; if (i!=j){ swap(a[i],a[j]); cnt++; } //向右 for (int k = 1,l = i;k <= n;k++,l++){ if (l>n) l = 1; v[k] = a[l]; } int temp1 = get_step(); //向左 for (int k = 1,l = i;k <= n;k++,l--){ if (l<=0) l = n; v[k] = a[l]; } int temp2 = get_step(); if (ans==-1){ ans = cnt+min(temp1,temp2); }else{ ans = min(ans,cnt+min(temp1,temp2)); } if (i!=j) swap(a[i],a[j]); } cout << ans << endl; } return 0; }
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