我的思路是 f[i][j] 表示考虑前 i 个人, 最后一队为第 j 队时最少的出列人数

f[i][j] = min{f[i-c][k]} + c-(s[i][j]-s[i-c][j])

其中, s[i][j] 表示前 i 个人中, 属于第 j 队的人数. c=s[n][j]

答案为 min{f[n][j]}

60 分. 说明有小 BUG 或者哪些地方没考虑到. 但我百思不得其解, 所以恳求路过大佬施与帮助.

#include <bits/stdc++.h>
#define REP(i, a, b) for(register int i = (a); i <=(b); ++i)
using namespace std;

ll read(){
    ll x=0; int f=0, c=getchar();
    for(; c<'0' || '9'<c; c=getchar()) f |= c=='-';
    for(; '0'<=c&&c<='9'; c=getchar()) x = (x<<1)+(x<<3)+int(c^'0');
    return f ?-x :x;
}

const int N = 1e5+9, M = 21;
const int INF = 0x3f3f3f3f;
int n, m;
int a[N], s[N][M], f[N][M], g[N];

int solve(){
    REP(i, 1, n){
        g[i] = INF;
        REP(j, 1, m){
            f[i][j] = INF;
            int c = s[n][j];
            if(i<c) continue;
            f[i][j] = g[i-c] + c-(s[i][j]-s[i-c][j]);
            g[i] = min(g[i], f[i][j]);
        }
    }
    int ans = INF;
    REP(i, 1, m) ans = min(ans, f[n][i]);
    return ans;
}

int main(){
    n = read(), m = read();
    REP(i, 1, n){
        a[i] = read();
        REP(j, 1, m){
            s[i][j] = s[i-1][j];
        }
        s[i][a[i]]++;
    }
    printf("%d\n", solve());
    return 0;
}