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=105;

ll rd(){
    ll x=0;char c=getchar();int neg=1;
    while(c<'0'||c>'9'){if(c=='-') neg=-1;c=getchar();}
    while(c>='0'&&c<='9') x=x*10+c-'0',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=1;i<=N;i++) num[i]=rd();
    for(i=1;i<=N;i++){
        for(j=i;j<=N;j++) dif[i][j]=dif[i][j-1]+(num[j]!=j-i);
    }
    memset(f,127,sizeof(f));f[0][0]=0;
    for(i=1;i<=N;i++){
        for(j=1;j<=i;j++){
            for(k=0;k<i;k++){
                f[i][j]=min(f[i][j],f[k][j-1]+dif[k+1][i]);
            }
        }
    }for(i=1;i<=N;i++) printf("%d\n",f[N][i]);
    return 0;
}