sgu482 Impudent Thief (动态规划)

设dp[i][j]为最后一个放i，周长为j所拿走的木板总高度

dC(i, j)表示i与j放在一起，周长的改变量

这样dp方程如下

dp[k][j + dC(i, k)] = max{dp[i][j] - h[k]} , k > i

时间复杂度为O(n^2*C)，即O(n^3*h)

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;

const int MAXN = 55;
const int MAXC = 5210;
const int INF = 0x3f3f3f3f;
int h[MAXN], sum[MAXN], dp[MAXN][MAXC], fat[MAXN][MAXC];
bool vis[MAXN];
vector<int> ansV;

inline int dC(int u, int v)
{
	int res = 2;
	if (h[v] > h[u]) res += 2 * (h[v] - h[u]);
	return res;
} 

int main()
{
	//input
	h[0] = sum[0] = 0;
	int sumC = 0;
	ansV.clear();
	memset(vis, false, sizeof(vis));
	int n; scanf("%d", &n);
	for (int i = 1; i <= n; i++)
	{
		scanf("%d", &h[i]);
		sum[i] = sum[i-1] + h[i];
		sumC += dC(i - 1, i);
	}
	int mid = (sumC + 1) / 2;
	
	//dp initialize
	memset(dp, 0, sizeof(dp));
	memset(fat, 0, sizeof(fat));
	for (int i = 1; i <= n; i++) dp[i][2*h[i]+2] = sum[n] - h[i];
	
	//dp
	for (int i = 1; i <= n; i++)
	for (int j = 0; j < MAXC; j++)
	{
		if (dp[i][j])
		{
			for (int k = i + 1; k <= n; k++)
			{
				if (dp[i][j] - h[k] > dp[k][j + dC(i, k)])
				{
					dp[k][j + dC(i, k)] = dp[i][j] - h[k];
					fat[k][j + dC(i, k)] = i;
				}
			}
		}
	}
	
	//get the biggest value
	int ans = 0, ansi = 0, ansj = 0;
	for (int i = 1; i <= n; i++)
	for (int j = mid; j < MAXC; j++)
	{
		if (dp[i][j] > ans)
		{
			ans = dp[i][j];
			ansi = i;
			ansj = j;
		}
	}
	
	//output
	printf("%d\n", ans);
	if (ans == 0)
	{
		printf("0\n");
		return 0;
	}
	int cnt = 0;
	while (ansi)
	{
		vis[ansi] = true; cnt++;
		int ti = fat[ansi][ansj];
		int tj = ansj - dC(ti, ansi);
		ansi = ti;
		ansj = tj;
	}
	printf("%d\n", n - cnt);
	bool flag = false;
	for (int i = 1; i <= n; i++)
	{
		if (!vis[i])
		{
			flag ? printf(" ") : flag = true;
			printf("%d", i);
		}
	}
	printf("\n");
	
	return 0;
}