任重而道遠

題目描述

In the good old days, Farmer John served a boring cuisine comprising but a single type of cow food to his N (1 <= N <= 40000) prize dairy cows. Times change. Today he serves the herd a total of M (1 <= M <= N) different types of food (conveniently numbered 1..M).

The cows are picky. Cow i has a single food preference P_i (1 <= P_i <= M) and will eat only that favorite food.

Each day at feeding time FJ converts the barn into a tastefully lit cafeteria. The cows line up outside to enter the cafeteria in order of their previously-mentioned convenient index number.

Unfortunately, with so many types of food, cleaning up afterwards is very time-consuming. If Farmer John is serving K different types of food, it takes him K*K units of time to clean the barn.

To save time, FJ serves the cows in contiguous groups from the line. After each group, he cleans up the barn and sets out the food for the next group (of course, he only sets out food that cows in the any given group will eat). Determine the minimum amount of total time FJ must spend cleaning the barn. Each group consists of the next contiguous group of cows from the line; each cow belongs to exactly one group; and the barn must be cleaned up after every group, including the last one.

輸入輸出格式

輸入格式：

* Line 1: Two space-separated integers: N and M

* Lines 2..N+1: Line i+1 contains a single integer: P_i

輸出格式：

* Line 1: A single integer: the minimum amount of time FJ must spend cleaning the barn.

輸入輸出樣例

輸入樣例#1： 複製

13 4 
1 
2 
1 
3 
2 
2 
3 
4 
3 
4 
3 
1 
4 

輸出樣例#1： 複製

11 

說明

There are four types of food and thirteen cows in line. The first cow prefers type 1, the second type 2, the third type 1, etc.

The first four groups contain one cow each. The fifth group contains two cows who prefer food #2 (requiring one unit of time). The sixth group contains cows preferring foods 3, 4, 3, 4, 3 (and requires four units of time to clean). The last two groups contain one cow each. The total time is 11.

AC程式碼：

#pragma optimize GCC ("O2")
#include<bits/stdc++.h>
using namespace std;

const int N = 4e4 + 5;
int p[N], last[N], pre[N], pos[N], nxt[N], cnt[N], dp[N];

int read () {
    int x = 0, f = 0; char c = getchar ();
    while (!isdigit (c)) f |= (c == '-'), c = getchar ();
    while (isdigit (c)) x = x * 10 + c - '0', c = getchar ();
    return f ? -x : x;
}

int main () {
    int n = read (), m = read (), blo = sqrt (n + 0.5);
    for (int i = 1; i <= n; i++) {
        p[i] = read ();
        pre[i] = last[p[i]]; nxt[last[p[i]]] = i;
        last[p[i]] = i; nxt[i] = n + 1;
    }
    memset (dp, 0x7f, sizeof (dp));
    dp[0] = 0;
    for (int i = 1; i <= blo; i++) pos[i] = 1;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= blo; j++) {
            if (pre[i] < pos[j]) cnt[j]++;
            if (cnt[j] > j) {
                while (nxt[pos[j]] < i) pos[j]++;
                pos[j]++, cnt[j]--;
            }
            dp[i] = min (dp[i], dp[pos[j] - 1] + j * j);
        }
    return !printf ("%d", dp[n]);
}