$\textit{\textbf 正解}$：

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洛谷题目传送器

拿出的书可以放在任意位置, 所以可以先放在一边, 到最后处理。

设 $F[i,j,k,s]$，即表示处理到第 $i$ 本书，拿出了 $j$ 本书，书架上最后一本书高度为 $k$，书架上的书高度状态为 $s$ 时的混乱值。

策略有两个：

1. 将 $i$ 拿出，$F[i,j,k,s]= \min (F[i-1,j-1,,k,s])$。

2. 不将 $i$ 拿出， $F[i,j,k,s]= \min (F[i-1.j,p,s \oplus (1<<h_i-1)]+[p \ne k])$。

再考虑统计答案， $Answer=\min (F[N,0 \to K,k,s]+ cnt [s \oplus U])$。

其中 $cnt[x]$ 表示 $x$ 二进制 $1$ 的个数。

按以上方式更新状态时间复杂度 $O(N^2*K*2^8)$ ，但若使用状态去更新状态, 时间复杂度 $O(N*K^2*2^9)$ 。

CODE：

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#include<bits/stdc++.h>
#define reg register
const int maxn = 105;
int read(){
    char c;
    int s = 0, flag = 1;
    while((c=getchar()) && !isdigit(c))
        if(c == '-'){ flag = -1, c = getchar(); break ; }
    while(isdigit(c)) s = s*10 + c-'0', c = getchar();
    return s * flag;
}
int N;
int K;
int num;
int Test_cnt;
int h[maxn];
int Mp[maxn];
int cnt[260];
int F[maxn][maxn][10][260];
int main(){
    for(reg int i = 1; i < 260; i ++) cnt[i] = cnt[i ^ (i & -i)] + 1;
    while((N = read()) && (K = read())){ 
        memset(Mp, 0, sizeof Mp); num = 0;
    for(reg int i = 1; i <= N; i ++){ 
        h[i] = read()-24;
            if(!Mp[h[i]]) Mp[h[i]] = ++ num;
            h[i] = Mp[h[i]];
    }
	for(reg int i = 0; i <= N; i ++)
        for(reg int j = 0; j <= K; j ++)
            for(reg int k = 0; k <= num; k ++)
                for(reg int s = 0; s < (1<<num); s ++) F[i][j][k][s] = 0x3f3f3f3f;
                F[0][0][0][0] = 0;
            		for(reg int i = 1; i <= N; i ++)
                        for(reg int j = 0; j <= K; j ++)
                            for(reg int k = 0; k <= num; k ++)
                                    for(reg int s = 0; s < (1<<num); s ++){
                                        if(j >= 1) F[i][j][k][s] = std::min(F[i][j][k][s], F[i-1][j-1][k][s]);
                                        if((s & (1<<h[i]-1)) && k == h[i]){
                                            int &t = F[i][j][k][s];
                                                for(reg int p = 0; p <= num; p ++){
                                                    t = std::min(t, F[i-1][j][p][s] + (p != k));
                                                    t = std::min(t, F[i-1][j][p][s ^ (1<<k-1)] + (p != k));
                                                    }
                                               	}
    										}
    int Ans = N, U = (1<<num)-1;
        for(reg int j = 0; j <= K; j ++)
            for(reg int i = 0; i <= num; i ++)
                for(reg int s = 0; s < (1<<num); s ++) Ans = std::min(Ans, F[N][j][i][s] + cnt[s ^ U]);
    printf("Case %d: %d\n\n", ++ Test_cnt, Ans);
    	}
    return 0;
}

整活:万行代码皆可压缩

#include<bits/stdc++.h>
#define reg register
const int maxn = 105;int read(){char c;int s = 0, flag = 1;while((c=getchar()) && !isdigit(c))if(c == '-'){ flag = -1, c = getchar(); break ; }while(isdigit(c)) s = s*10 + c-'0', c = getchar();return s * flag;}int N;int K;int num;int Test_cnt;int h[maxn];int Mp[maxn];int cnt[260];int F[maxn][maxn][10][260];int main(){for(reg int i = 1; i < 260; i ++) cnt[i] = cnt[i ^ (i & -i)] + 1;while((N = read()) && (K = read())){ memset(Mp, 0, sizeof Mp); num = 0;for(reg int i = 1; i <= N; i ++){h[i] = read()-24;if(!Mp[h[i]]) Mp[h[i]] = ++ num;h[i] = Mp[h[i]];}for(reg int i = 0; i <= N; i ++)for(reg int j = 0; j <= K; j ++)for(reg int k = 0; k <= num; k ++)for(reg int s = 0; s < (1<<num); s ++) F[i][j][k][s] = 0x3f3f3f3f;F[0][0][0][0] = 0;for(reg int i = 1; i <= N; i ++)for(reg int j = 0; j <= K; j ++)for(reg int k = 0; k <= num; k ++)for(reg int s = 0; s < (1<<num); s ++){if(j >= 1) F[i][j][k][s] = std::min(F[i][j][k][s], F[i-1][j-1][k][s]);if((s & (1<<h[i]-1)) && k == h[i]){int &t = F[i][j][k][s];for(reg int p = 0; p <= num; p ++){t = std::min(t, F[i-1][j][p][s] + (p != k));t = std::min(t, F[i-1][j][p][s ^ (1<<k-1)] + (p != k));}}}int Ans = N, U = (1<<num)-1;for(reg int j = 0; j <= K; j ++)for(reg int i = 0; i <= num; i ++)for(reg int s = 0; s < (1<<num); s ++) Ans = std::min(Ans, F[N][j][i][s] + cnt[s ^ U]);printf("Case %d: %d\n\n", ++ Test_cnt, Ans);}return 0;}