P1284 三角形牧场

链接

https://www.luogu.com.cn/problem/P1284

思路

一个dp。dp状态：前k根，两边长为i，j的时候的可行性。
dp[k][i][j] = dp[k-1][i-a[k]][j] || dp[k-1][i][j-a[k]]|| f[k-1][i][j];
如果可行并且满足构建三角形就直接取面积的max

代码

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define tin long long
#define itn long long
#define IOS ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
void solve();

const int N = 42;
int a[N];
int n;
int sumint;
int dp[2][N * N / 2][N * N / 2];
bool check(int a, int b, int c)
{
	if (a + b <= c)return false;
	if (b+c <= a)return false;
	if (a + c <= b)return false;
	return true;
}
double cal(int a, int b, int c)
{
	double p = ((double)a + b + c) / 2;
	return sqrt(p * (p - a) * (p - b) * (p - c));
}
void solve()
{
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> a[i]; sumint += a[i];
	}
	int now = 0, old = 1;
	dp[now][0][0] = 1;
	for (int k = 1; k <= n; k++)
	{
		swap(now, old);
		for(int i=sumint/2;i>=0;i--)
			for (itn j = 0; j <= sumint / 2;j++)
			{
				dp[now][i][j] = dp[old][i][j];
				if (i >= a[k] and dp[old][i - a[k]][j])dp[now][i][j] = 1;
				if (j >= a[k] and dp[old][i][j - a[k]])dp[now][i][j] = 1;
				
			}
	}
	double ans = -1;
	for(int i=sumint/2;i-->=0;)
		for (int j = sumint / 2; j-- >= 0;)
		{
			if (check(i, j, sumint - i - j) and dp[now][i][j])
				ans = max(ans, cal(i, j, sumint - i - j));
		}

	if (ans == (double)-1) { cout << -1; return; }
	ans *= 100;
	cout << (int)ans;
}


signed main()
{
	IOS;

	solve();
	return 0;
}