luogu4267 TamingtheHerd (dp)

设f[i][j]为让前i天发生j次暴动需要改变的最少的值

则f[i][j]=min{f[k][j-1]+(x[k+1]!=0)+(x[k+2]!=1)+...+(x[i]!=(i-k-1))}

$O(n^3)$直接做就好了

 #include<bits/stdc++.h>
 #define pa pair<int,int>
 #define ll long long
 using namespace std;
 const int maxn=;

 ll rd(){
     ll x=;char c=getchar();int neg=;
     while(c<''||c>''){if(c=='-') neg=-;c=getchar();}
     while(c>=''&&c<='') x=x*+c-'',c=getchar();
     return x*neg;
 }

 int N,num[maxn],f[maxn][maxn],dif[maxn][maxn];

 int main(){
     int i,j,k;
     N=rd();for(i=;i<=N;i++) num[i]=rd();
     for(i=;i<=N;i++){
         for(j=i;j<=N;j++) dif[i][j]=dif[i][j-]+(num[j]!=j-i);
     }
     memset(f,,sizeof(f));f[][]=;
     for(i=;i<=N;i++){
         for(j=;j<=i;j++){
             for(k=;k<i;k++){
                 f[i][j]=min(f[i][j],f[k][j-]+dif[k+][i]);
             }
         }
     }for(i=;i<=N;i++) printf("%d\n",f[N][i]);
     return ;
 }