luogu1005 矩阵取数游戏

最新推荐文章于 2025-05-09 13:03:24 发布

转载最新推荐文章于 2025-05-09 13:03:24 发布 · 89 阅读

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原文链接：http://www.cnblogs.com/headboy2002/p/9370703.html

本文介绍了一种求解矩阵中取数最大价值的问题。通过递归深度优先搜索结合大整数运算，实现对每行两端数值的选择，以获得整体的最大价值。

题目大意

　　一个矩阵，每次从每一行的行首或行尾取一个数，每一行的价值为取的数*2^当前取数的次数，每一次的价值为每一行的价值的和。求得到的价值的最大值。

思路

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

const int MAX_ROW = 100, MAX_COL = 100;
int A[MAX_ROW][MAX_COL];
int TotRow, TotCol;

struct BigInt
{
private:
	static const int MAX_N = 100, BASE = 10000, CARRY = 4;
	int A[MAX_N];
	int Len;

public:
	void Print()
	{
		printf("%d", A[Len]);
		for (int i = Len - 1; i >= 0; i--)
			printf("%0*d", CARRY, A[i]);
		printf("\n");
	}

	void Clear()
	{
		memset(A, 0, sizeof(A));
		Len = 0;
	}

	void Set(int x)
	{
		Clear();
		while (x)
		{
			A[Len++] = x % BASE;
			x /= BASE;
		}
		while (Len > 0 && A[Len] == 0)
			Len--;
	}

	BigInt(int x)
	{
		Set(x);
	}

	BigInt()
	{
		Set(0);
	}

	BigInt operator =(const BigInt& a)
	{
		memcpy(A, a.A, sizeof(A));
		Len = a.Len;
		return *this;
	}

	BigInt operator *=(const BigInt& a)
	{
		BigInt b = *this;
		Clear();
		Len = a.Len + b.Len;
		for (int i = 0; i <= a.Len; i++)
			for (int j = 0; j <= b.Len; j++)
			{
				A[i + j] += a.A[i] * b.A[j];
				A[i + j + 1] += A[i + j] / BASE;
				A[i + j] %= BASE;
			}
		if (A[Len + 1])
			Len++;
		return *this;
	}

	BigInt operator *(const BigInt& a)
	{
		BigInt ans = *this;
		ans *= a;
		return ans;
	}

	BigInt operator +=(const BigInt& a)
	{
		Len = max(Len, a.Len);
		for (int i = 0; i <= Len; i++)
		{
			A[i] += a.A[i];
			A[i + 1] += A[i] / BASE;
			A[i] %= BASE;
		}
		if (A[Len + 1])
			Len++;
		return *this;
	}

	BigInt operator +(const BigInt& a)
	{
		BigInt ans = *this;
		ans += a;
		return ans;
	}

	bool operator <(const BigInt& a) const
	{
		if (Len != a.Len)
			return Len < a.Len;
		for (int i = Len; i >= 0; i--)
			if (A[i] != a.A[i])
				return A[i] < a.A[i];
		return true;
	}

	bool Is0()
	{
		return Len == 0 && A[Len] == 0;
	}
}F[MAX_COL][MAX_COL], Pow2[MAX_COL];
bool Vis[MAX_COL][MAX_COL];

void InitPow2(int n)
{
	Pow2[0] = 1;
	for (int i = 1; i <= n; i++)
		Pow2[i] = Pow2[i - 1] * 2;
}

BigInt Dfs(int row, int l, int r)
{
	if (l > r)
		return 0;
	if (Vis[l][r])
		return F[l][r];
	Vis[l][r] = true;
	BigInt a = Dfs(row, l + 1, r) + Pow2[TotCol - r + l] * A[row][l];
	BigInt b = Dfs(row, l, r - 1) + Pow2[TotCol - r + l] * A[row][r];
	return F[l][r] = a < b ? b : a;
}

BigInt CalRow(int row)
{
	memset(Vis, false, sizeof(Vis));
	return Dfs(row, 1, TotCol);
}

int main()
{
	scanf("%d%d", &TotRow, &TotCol);
	InitPow2(TotCol);
	for (int i = 1; i <= TotRow; i++)
		for (int j = 1; j <= TotCol; j++)
			scanf("%d", &A[i][j]);
	static BigInt ans(0);
	for (int row = 1; row <= TotRow; row++)
		ans += CalRow(row);
	ans.Print();
	return 0;
}

转载于:https://www.cnblogs.com/headboy2002/p/9370703.html