DP(优化专题-四边形不等式优化一)

题意: 有若干堆石子围成一圈儿, 每合并两堆石子, 就对答案贡献了这两堆石子的重量, 现询问答案的最大值与最小值.

>> face <<

状态: $dpmin[l][r]\to$该区间内的最小收益,$dpmax[l][r]\to$该区间内最大收益

目标:$dpmin[1][n]\&dpmax[1][n]$

边界: 第一次转移的贡献全由石堆提供. 不需要额外提供边界(记忆化搜索),针对区间最小,首先要memset(dpmin, $\infty$, sizeof(dpmin)), 然后确保$dpmin[i][i]=0,i\in [1,2\ast n]$

合法判断: 本题无

转移方程: 区间dp, 枚举未知, 让所有能转移到该未知的状态来转移

$$\{\begin{array}{l}dpmin[l][r]=min(dpmin[l][r],dpmin[l][k]+dpmin[k+1][r])+\sum _{i=l}^{i\le r}a[i];\\ dpmax[l][r]=max(dpmax[l][r],dpmax[l][k]+dpmax[k+1][r])+\sum _{i=l}^{i\le r}a[i];\end{array}$$

attention: 圆环链化 (化为长度为2*n的链)

双倍经验:

Strategy:区间dp(用四边形不等式优化)

先贴一波大佬的博客

• 若想用四边形不等式优化必须满足 决策单调性, 就本题而言, 针对dpmin[i][j]都有某最K值使得该最小值被取到, 而且, 该最小值是和决策区间[i][j]有非严格单调关系的情况, 那么我们就可以用一个二维数组存下来, 然后循环找k的时候可以优化一层

首先给出正确的$O(n^3)$代码

#include 
#include 
#define _rep(i, a, b) for (int i = (a); i <= (b); ++i)
#define _rev(i, a, b) for (int i = (a); i >= (b); --i)
#define _for(i, a, b) for (int i = (a); i < (b); ++i)
#define _rof(i, a, b) for (int i = (a); i > (b); --i)
#define ll long long
#define db double
#define oo 0x3f3f3f3f
#define eps 0.00001
#define all(x) x.begin(), x.end()
#define met(a, b) memset(a, b, sizeof(a))
#define id(x) ((x + 8))
#define what_is(x) cerr << #x << " is " << x << endl
#define lowbit(x) x &(-x)
using namespace std;
const int maxn = 2e3 + 9;
const int mod = 1e6 + 3;
int dpmin[maxn][maxn], n, a[maxn], dpmax[maxn][maxn], sum[maxn];
int main()
{
	ios::sync_with_stdio(0);
	int n;
	cin >> n;
	met(dpmin, oo);
	_rep(i, 1, n)
	{
		cin >> a[i];
		a[i + n] = a[i];
		dpmin[i][i] = dpmin[i + n][i + n] = 0;
	}
	_rep(i, 1, 2*n){
		sum[i] = sum[i-1] + a[i];
	}
	_rep(len, 2, n)
	{
		_rep(l, 1, 2 * n - len + 1)
		{
			int r = l + len - 1;
			_rep(k, l, r-1){
				dpmax[l][r] = max(dpmax[l][k] + dpmax[k+1][r] + sum[r] - sum[l-1], dpmax[l][r]);
				dpmin[l][r] = min(dpmin[l][k] + dpmin[k+1][r] + sum[r] - sum[l-1], dpmin[l][r]);
			}
		}
	}

	int min_val = oo, max_val = 0;

	_rep(i, 1, n)
	{
		min_val = min(dpmin[i][i + n - 1], min_val);
		max_val = max(max_val, dpmax[i][i + n - 1]);
	}
	cout << min_val << endl
		 << max_val << endl;
}

然后打表判断决策单调性

#include
#include
#include
#include
#include
#define _rep(i, a, b) for (int i = (a); i <= (b); ++i)
#define _rev(i, a, b) for (int i = (a); i >= (b); --i)
#define _for(i, a, b) for (int i = (a); i < (b); ++i)
#define _rof(i, a, b) for (int i = (a); i > (b); --i)
#define ll long long
#define db double
#define oo 0x3f3f3f3f
#define eps 0.00001
#define all(x) x.begin(), x.end()
#define met(a, b) memset(a, b, sizeof(a))
#define id(x) ((x + 8))
#define what_is(x) cerr << #x << " is " << x << endl
#define lowbit(x) x &(-x)
using namespace std;
const int maxn = 2e3 + 9;
const int mod = 1e6 + 3;
int dpmin[maxn][maxn], n, a[maxn], dpmax[maxn][maxn], sum[maxn], kma[maxn][maxn], kmi[maxn][maxn];
int main()
{
	ios::sync_with_stdio(0);
	int n;
	cin >> n;
	met(dpmin, oo);
	_rep(i, 1, n)
	{
		cin >> a[i];
		a[i + n] = a[i];
		dpmin[i][i] = dpmin[i + n][i + n] = 0;
	}
	_rep(i, 1, 2 * n) {
		sum[i] = sum[i - 1] + a[i];
		kmi[i][i] = kma[i][i] = i;
	}
	_rep(len, 2, n)
	{
		_rep(l, 1, 2 * n - len + 1)
		{
			int r = l + len - 1;
			_rep(k, l, r - 1) {
				/*dpmax[l][r] = max(dpmax[l][k] + dpmax[k + 1][r] + sum[r] - sum[l - 1], dpmax[l][r]);
				dpmin[l][r] = min(dpmin[l][k] + dpmin[k + 1][r] + sum[r] - sum[l - 1], dpmin[l][r]);*/
				if (dpmax[l][k] + dpmax[k + 1][r] + sum[r] - sum[l - 1] > dpmax[l][r]) {
					dpmax[l][r] = dpmax[l][k] + dpmax[k + 1][r] + sum[r] - sum[l - 1];
					kma[l][r] = k;
				}
				if (dpmin[l][k] + dpmin[k + 1][r] + sum[r] - sum[l - 1] < dpmin[l][r]) {
					dpmin[l][r] = dpmin[l][k] + dpmin[k + 1][r] + sum[r] - sum[l - 1];
					kmi[l][r] = k;
				}
			}
		}
	}
	_rep(i, 1, n + 1) {
		_rep(j, i+1, i + n-1) {
			assert(kma[i][j - 1] <= kma[i][j] && kma[i][j] <= kma[i + 1][j]);
			//assert(kmi[i][j - 1] <= kmi[i][j] && kmi[i][j] <= kmi[i + 1][j]);
		}
	}
}

发现只有kmi满足决策单调性, 而kma不满足

但是对于dpmax有一个性质, 对于某dpmax[l][r], 只有k = l或者k = r-1时最大

其实最小也有性质, 当k在l和r中间的时候也能取最小

优化后代码

#include 
#include 
#define _rep(i, a, b) for (int i = (a); i <= (b); ++i)
#define _rev(i, a, b) for (int i = (a); i >= (b); --i)
#define _for(i, a, b) for (int i = (a); i < (b); ++i)
#define _rof(i, a, b) for (int i = (a); i > (b); --i)
#define ll long long
#define db double
#define oo 0x3f3f3f3f
#define eps 0.00001
#define all(x) x.begin(), x.end()
#define met(a, b) memset(a, b, sizeof(a))
#define id(x) ((x + 8))
#define what_is(x) cerr << #x << " is " << x << endl
#define lowbit(x) x &(-x)
using namespace std;
const int maxn = 2e3 + 9;
const int mod = 1e6 + 3;
int dpmin[maxn][maxn], n, a[maxn], dpmax[maxn][maxn], sum[maxn], s[maxn][maxn] ; //s[i][j]存储dpmin[i][j]的最优解k
int main()
{
	ios::sync_with_stdio(0);
	int n;
	cin >> n;
	met(dpmin, oo);
	_rep(i, 1, n)
	{
		cin >> a[i];
		a[i + n] = a[i];
		dpmin[i][i] = dpmin[i + n][i + n] = 0;
	}
	_rep(i, 1, 2 * n)
	{
		sum[i] = sum[i - 1] + a[i];
		s[i][i] = i;
	}
	_rep(len, 2, n)
	{
		_rep(l, 1, 2 * n - len + 1)
		{
			int r = l + len - 1;
			dpmax[l][r] = max(dpmax[l+1][r]+dpmax[l][l], dpmax[l][r-1]+dpmax[r][r]) + sum[r] - sum[l-1];
			int K, tmp = oo;
			_rep(k, s[l][r-1], s[l+1][r]){
				int tt = dpmin[l][k] + dpmin[k+1][r] + sum[r] - sum[l-1];
				if(tt < tmp){
					tmp = tt;
					K = k;
				}
			}
			s[l][r] = K;
			dpmin[l][r] = tmp;
		}
	}
	

	int min_val = oo, max_val = 0;

	_rep(i, 1, n)
	{
		min_val = min(dpmin[i][i + n - 1], min_val);
		max_val = max(max_val, dpmax[i][i + n - 1]);
	}
	cout << min_val << endl
		 << max_val << endl;
}