DP Classic Bzoj 1584: [Usaco2009 mar]cleaning up cleaning

Bzoj 1584: [Usaco2009 mar]cleaning up cleaning time limit:10 Sec Memory limit:64 MB
submit:419 solved:278
Description

There are N cows, each of which has a label pi,1 <= Pi <= M <= N <= 40000. Now farmer John is going to divide these cows into segments, defining the non-crab degree for each paragraph: if there are k different numbers in this paragraph, then the degree of the river Crab is k*k. The total non-crab degree is the sum of all sections of the non-crab degree.

Input

First line: two integers n,m

2nd.. N+1 line: n integers represent the number of each cow

Output

An integer representing the minimum non-crab degree

Sample Input13 4
1
2
1
3
2
2
3
4
3
4
3
1
4
Sample Output One

1 /*because this topic grouping is unlimited, so we can not put the DP equation as a state2 Positive Solution: The worst case of each digit in a paragraph, ans=n, so if there is a range of different number of <=SQRT (n), otherwise the result must not be optimal. 3 nsqrt (N) Method: Maintain B[j],c[j],f[j],pre[j] array. 4 B[j] Indicates that the B[J]+1...I has the leftmost end of the interval of a different number of J. 5 so you know f[i]=min{f[i],f[b[j]]+j*j}; So the complexity of time falls down.6 how to maintain the B[j] array, when I move backward one bit, pre[a[i]] record A[i] where is the last position to appear? 7 well: After i++, pre[a[i]]<=b[j], description b[j]+1 ... to the number of different numbers of this sequence is j+1, we use c[j] to record the situation, by updating pre[a[i]], and always maintain c[j]==j;8 then B[j] is still in line with test instructions. 9 Maintenance C[j] from the beginning of the b[j]+1 to delete data, delete when judged if pre[a[t]]>t, it is deleted the same number, for the final harmony value has no effect, so also to delete the numberTen know pre[a[t]]<=t, delete a[t], by the way update B[j] location One */ A #defineN 40100 -#include <iostream> - using namespacestd; the#include <cstdio> -#include <cmath> -#include <cstring> - intF[n],b[n],c[n],pre[n],a[n]; + intn,m; - voidinput () + { Ascanf"%d%d",&n,&m); at      for(intI=1; i<=n;++i) -scanf"%d",&a[i]); -memset (pre,-1,sizeof(pre));/*do not forget to set to-1, because the following will be compared with the initial value of B[j]==0*/ -Memset (F,127,sizeof(f)); - } - voidChuli () in { -     intSQRTN=SQRT (n+0.5); tof[0]=0;/*initialization, the first 0 number of disharmony value is 0,*/ +      for(intI=1; i<=n;++i) -     { the          for(intj=1; j<=sqrtn;++j) *         { $             if(pre[a[i]]<=B[j])Panax Notoginsengc[j]++;/*statistics on the number of newly added are not in line with the requirements*/ -         } thePre[a[i]]=i;/*Update Pre*/ +          for(intj=1; j<=sqrtn;++j) A         { the             if(C[J]&GT;J)/*Delete number, shorten sequence*/ +             { -                 intt=b[j]+1; $                  while(pre[a[t]]>t) + +T; $b[j]=t;c[j]--; -             } -         } the          for(intj=1; j<=sqrtn;++j) -F[i]=min (F[I],F[B[J]]+J*J);/*Update F*/Wuyi     } the } - intMain () Wu { - input (); About Chuli (); $cout<<f[n]<<Endl; -     return 0; - } - /*since this topic is not divided into the number of States, then you can consider, the length of the dividing sequence*/

DP Classic Bzoj 1584: [Usaco2009 mar]cleaning up cleaning