将序列储存两遍,用数组代替环,然后枚举起点,每次都把当前数交换过来,即可得到答案。
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| #include <cstdio> #include <algorithm> #include <cstring> using namespace std;
const int maxn = 1010; const int inf = 0x3f3f3f3f; int a[maxn], b[maxn];
int main() { int n; while(~scanf("%d", &n) && n) { for(int i = 1; i <= n; ++i) scanf("%d", &a[i]), a[n + i] = a[i]; int ans = inf, cur = 0; for(int i = 1; i <= n; ++i) { memcpy(b, a, sizeof(a)); cur = 0; for(int j = 1; j <= n && cur < ans; ++j) { int k = i + j - 1; if(b[k] != j) { ++cur; int pos = k + 1; while(pos < i + n) { if(b[pos] == j) {swap(b[k], b[pos]); break;} ++pos; } } } ans = min(ans, cur); memcpy(b, a, sizeof(a)); cur = 0; for(int j = n; j >= 1 && cur < ans; --j) { int k = i + n - j; if(b[k] != j) { ++cur; int pos = k + 1; while(pos < i + n){ if(b[pos] == j) {swap(b[k], b[pos]); break;} ++pos; } } } ans = min(ans, cur); } printf("%d\n", ans); } return 0; }
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