CF448C [Painting Fence]递归分治

题目链接：http://codeforces.com/problemset/problem/448/C

题目大意：用宽度为1的刷子刷墙，墙是一长条一长条并在一起的。梳子可以一横或一竖一刷到底。求刷完整面墙最少要几次。

关键思路：很有意思，我们会发现，一座墙，从最短的那根往下的矩形区域，全刷完后会形成多座“小山”，这和最先要刷的墙形状是一样的。所以问题就可以进行分治了。

代码如下：

//图像类比
#include<iostream>
#include<algorithm>
using namespace std;
int fence[10000] = { 0 };

int solve(int s, int n) {
	int minh = fence[s];
	int times = 0;
	if (n == 1)return 1;
	for (int i = 0; i < n; i++)
		fence[s + i] <= minh ? minh = fence[s + i] : minh;
	for (int i = 0; i < n; i++)
		fence[s + i] -= minh;
	times += minh;
	int start = 0, end = 0;
	for (int i = 0; i <= n; i++) {
		if ((fence[s + i] != 0) && (fence[s + i - 1] == 0))start = s + i;
		if ((fence[s + i] == 0) && (fence[s + i - 1] != 0)) {
			end = s + i - 1;
			times += solve(start, end - start + 1);
			continue;
		}
	}

	return min(times, n);
}

int main() {
	int Total;
	cin >> Total;

	int i = 0;
	for (i = 0; i < Total; i++)
		cin >> fence[i];

	cout << solve(0, Total)<<endl;
	//system("Pause");
	return 0;

}