2020寒假集训排位赛 Snakes题解（区间dp）

专题链接：https://codeforces.com/group/5yyKg9gx7m/contest/269908/problem/B

思路

---

预先处理出【l，r】在只有一种网的前提下的浪费空间，再dp找出最小的浪费空间即可。（说起来简单

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<string>
#include<map>
#include<cstring>
#include<climits>
#include<vector>
#include<queue>
#include<stack>
#define lowbit(i) ((i)&(-(i)))
#define ll long long
using namespace std;
int N, K, ans;
int arr[401];
int g[401][401];//只调整一次的情况下代表i到j的最小浪费空间
int f[401][401];//前i条蛇调整j次的最小代价
int sum[401];
int main()
{
	cin >> N >> K;
	K++;
	for (int i = 1; i <= N; i++)
		scanf("%d", arr + i);
	for (int i = 1; i <= N; i++)//遍历长度
	{
		int maxh = 0;
		for (int j = i; j <= N; j++)
		{
			if (arr[j] > maxh)
			{
				g[i][j] = g[i][j - 1] + (j - i) * (arr[j] - maxh);
				maxh = arr[j];
			}
			else
				g[i][j] = g[i][j - 1] + maxh - arr[j];
		}
	}
	memset(f, 0x3f, sizeof f);
	f[0][0] = 0;
	for (int i = 1; i <= N; i++)//遍历长度
		for (int j = 1; j <= K; j++)//遍历改变次数
			for (int t = 0; t < i; t++)//遍历改变的位置
				f[i][j] = min(f[i][j], f[t][j - 1] + g[t + 1][i]);
	printf("%d\n", f[N][K]);

	return 0;
}