[Codeforces 448C]Painting Fence

Description

Bizon the Champion isn't just attentive, he also is very hardworking.

Bizon the Champion decided to paint his old fence his favorite color, orange. The fence is represented as n vertical planks, put in a row. Adjacent planks have no gap between them. The planks are numbered from the left to the right starting from one, the i-th plank has the width of 1 meter and the height of ai meters.

Bizon the Champion bought a brush in the shop, the brush's width is 1 meter. He can make vertical and horizontal strokes with the brush. During a stroke the brush's full surface must touch the fence at all the time (see the samples for the better understanding). What minimum number of strokes should Bizon the Champion do to fully paint the fence? Note that you are allowed to paint the same area of the fence multiple times.

Input

The first line contains integer n (1 ≤ n ≤ 5000) — the number of fence planks. The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

Print a single integer — the minimum number of strokes needed to paint the whole fence.

Sample Input

5
2 2 1 2 1

Sample Output

3

HINT

In the first sample you need to paint the fence in three strokes with the brush: the first stroke goes on height 1 horizontally along all the planks. The second stroke goes on height 2 horizontally and paints the first and second planks and the third stroke (it can be horizontal and vertical) finishes painting the fourth plank.

题解

我们在$[1,n]$区间内，如果要横着涂，那么必定要从最底下往上涂$min(h_1,h_2,...h_n)$次，那么又会将$[1,n]$的序列分成多个子局面。

对于每个子局面，我们分治，单独求解，求解函数递归实现。复杂度$O(n^2)$。

//It is made by Awson on 2017.10.9
#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define query QUERY
#define LL long long
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
using namespace std;
const int N = 5000;

int n, a[N+5];

int doit(int l, int r, int base) {
    if (l == r) return 1;
    int high = 2e9;
    for (int i = l; i <= r;  i++)
        high = Min(high, a[i]);
    int ans = high-base;
    for (int i = l; i <= r; i++)
        if (a[i] != high) {
            int j;
            for (j = i; j <= r && a[j+1] > high; j++);
            ans += doit(i, j, high);
            i = j+1;
    }
    return Min(ans, r-l+1);
}
void work() {
    scanf("%d", &n);
    for (int i = 1; i <= n ; i++) scanf("%d", &a[i]);
    printf("%d
", doit(1, n, 0));
}
int main() {
    work();
    return 0;
}