BZOJ1584 [Usaco2009 Mar]Cleaning Up 打扫卫生

令$f[i]$表示以i为结尾的答案最小值，则$f[i] = min \{f[j] + cnt[j + 1][i]^2\}_{1 \leq j < i}$，其中$cnt[j + 1][i]$表示$[j + 1, i]$内有几个不同的数

对于区间长度为$k$，则答案最大值就是$\sqrt{k}$，所以对于每个$i$我们其实只要枚举$\sqrt{i}$个值就好了

 /**************************************************************
     Problem: 1584
     User: rausen
     Language: C++
     Result: Accepted
     Time:112 ms
     Memory:1120 kb
 ****************************************************************/

 #include <cstdio>
 #include <algorithm>

 using namespace std;
 const int N = 4e4 + ;
 const int MXlen = ;

 int n, m;
 int a[N], f[N], seq[MXlen], now[MXlen];
 int st, mxlen, nowlen;

 inline int read();

 inline int sqr(int x) {
     return x * x;
 }

 int main() {
     int i, j;
     n = read(), m = read();
     for (i = ; i <= n; ++i) a[i] = read();
     for (mxlen = ; mxlen * mxlen <= n; ++mxlen);
     --mxlen;
     for (i = ; i <= n; ++i) {
         f[i] = i;
         for (st = , j = ; j <= nowlen; ++j)
             if (seq[j] == a[i]) {
                 st = j;
                 break;
             }
         if (!st) {
             if (nowlen != mxlen) {
                 seq[++nowlen] = a[i];
                 now[nowlen] = i;
             } else {
                 for (j = ; j < nowlen; ++j)
                     seq[j] = seq[j + ], now[j] = now[j + ];
                 seq[nowlen] = a[i], now[nowlen] = i;
             }
         } else {
             for (j = st; j < nowlen; ++j)
                 seq[j] = seq[j + ], now[j] = now[j + ];
             seq[nowlen] = a[i], now[nowlen] = i;
         }
         for (j = nowlen; j >= ; --j)
             f[i] = min(f[i], f[now[j - ]] + sqr(nowlen - j + ));
     }
     printf("%d\n", f[n]);
     return ;
 }

 inline int read() {
     static int x;
     static char ch;
     x = , ch = getchar();
     while (ch < '' || '' < ch)
         ch = getchar();
     while ('' <= ch && ch <= '') {
         x = x *  + ch - '';
         ch = getchar();
     }
     return x;
 }