uva 11212 UVa11212 Editing a Book

此题题目意思很简单：就是问最少的可以剪切和粘贴的次数。

但是实现起来确实是很麻烦，主要是思路不清晰

此题加深了我对迭代加深搜索的理解  同时其中剪枝是非常难以想到的。

借用刘汝佳先生书上所说：此题采用迭代加深搜索，下面是刘汝佳先生的代码

<span style="font-size: 13.3333339691162px; font-family: Arial, Helvetica, sans-serif;">// UVa11212 Editing a Book</span>

// Rujia Liu
// This implementation is not very fast, but easy to understand
#include<cstdio>
#include<cstring>
using namespace std;

const int maxn = 9;
int n, a[maxn];

bool is_sorted() {
  for(int i = 0; i < n-1; i++)
    if(a[i] >= a[i+1]) return false;
  return true;
}

// the number of integers with incorrect successor
int h() {
  int cnt = 0;
  for(int i = 0; i < n-1; i++)
    if(a[i]+1 != a[i+1]) cnt++;
  if(a[n-1] != n) cnt++;
  return cnt;
}

bool dfs(int d, int maxd) {
  if(d*3 + h() > maxd*3) return false;
  if(is_sorted()) return true;

  int b[maxn], olda[maxn];
  memcpy(olda, a, sizeof(a));
  for(int i = 0; i < n; i++)
   for(int j = i; j < n; j++) {

     // cut and paste  are <span style="font-family: Arial, Helvetica, sans-serif; font-size: 12px;">the difficult  part.</span>

     int cnt = 0;
     for(int k = 0; k < n; k++)
       if(k < i || k > j) b[cnt++] = a[k];

     // insert before position k
     for(int k = 0; k <= cnt; k++) {
       int cnt2 = 0;
       for(int p = 0; p < k; p++) a[cnt2++] = b[p];
       for(int p = i; p <= j; p++) a[cnt2++] = olda[p];
       for(int p = k; p < cnt; p++) a[cnt2++] = b[p];

       if(dfs(d+1, maxd)) return true;
       memcpy(a, olda, sizeof(a));
     }
   }
  return false;
}

int solve() {
  if(is_sorted()) return 0;
  int max_ans = 5; // after experiments, we found ans <= 5 for n <= 9
  for(int maxd = 1; maxd < max_ans; maxd++)
    if(dfs(0, maxd)) return maxd;
  return max_ans;
}

int main() {
  int kase = 0;
  while(scanf("%d", &n) == 1 && n) {
    for(int i = 0; i < n; i++) scanf("%d", &a[i]);
    printf("Case %d: %d\n", ++kase, solve());
  }
  return 0;
}