题目链接:Editing a Book
Description
给定一个(1)~(n)的全排列,你可以每次选择其中一段区间,将其剪切,并黏贴到任意位置。
问最少操作几次可以将这个排列变成 (1,2,...,n) 的形式。
多组数据。
数据范围 数据组数(le 50),(1le nle 9)
时间限制 (20 Sec)
Solution
我们考虑(IDA*),每一次操作最多改变3个数字的后继。
定义(h()=sum_{i=2}^{n} [num[i] != num[i-1]+1]),表示有多少个相邻对不满足顺序。
那么,如果(h()>3(maxd-d)),代表进行完所有操作,并且每次操作都修改3个数字的后继,都不能满足条件,则立刻返回。
爆搜即可。
复杂度 (O(玄学))
bzoj土豆机过于土豆,导致此代码本地能过,提交TLE。
其实还有更优秀的做法,复杂度是(O(n^6 log(n!)))的,具体见(Claris)的博客。
https://www.cnblogs.com/clrs97/p/6321773.html
Code
// Author: wlzhouzhuan
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define rint register int
#define rep(i, l, r) for (rint i = l; i <= r; i++)
#define per(i, l, r) for (rint i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
inline int read() {
int x = 0, neg = 1; char op = getchar();
while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
return neg * x;
}
inline void print(int x) {
if (x < 0) { putchar('-'); x = -x; }
if (x >= 10) print(x / 10);
putchar(x % 10 + '0');
}
const int N = 10;
int a[N], n, maxd, flag;
int h() {
int cnt = 0;
for (rint i = 2; i <= n; i++) {
if (a[i - 1] + 1 != a[i]) {
cnt++;
}
}
return cnt;
}
bool check() {
for (rint i = 1; i <= n; i++) {
if (a[i] != i) {
return 0;
}
}
return 1;
}
void dfs(int x) {
if (flag) return ;
if (check()) {
flag = 1;
return ;
}
if (h() > 3 * (maxd - x)) {
return ;
}
int b[10], c[10], len;
for (rint i = 1; i <= n; i++) b[i] = a[i];
for (rint le = 1; le <= n; le++) {
for (rint ri = le; ri <= n; ri++) {
len = 0;
for (rint i = 1; i <= n; i++) if (i < le || i > ri) c[++len] = b[i];
for (rint cut = 0; cut <= len; cut++) {
int cur = 0;
for (rint i = 1; i <= cut; i++) a[++cur] = c[i];
for (rint i = le; i <= ri; i++) a[++cur] = b[i];
for (rint i = cut + 1; i <= len; i++) a[++cur] = c[i];
dfs(x + 1);
}
}
}
}
int main() {
while (~scanf("%d", &n) && n) {
for (int i = 1; i <= n; i++) {
a[i] = read();
}
flag = 0;
for (maxd = 0; maxd <= 10; maxd++) {
dfs(0);
if (flag) break;
}
printf("%d
", maxd);
}
return 0;
}