超超超超级详细的多边形游戏问题分析（动态规划）
- 多边形游戏
- 问题简介
- 问题分析
- 超详细解题步骤
- Java代码实现
多边形游戏
问题简介
- 首先呢，介绍一下多边形游戏是个什么东东
- 多边形游戏是一个单人玩的游戏，开始时有一个由n个顶点构成的多边形。每个顶点被赋予一个整数值，每条边被赋予一个运算符“+”或“*”。所有边依次用整数从1到n编号。
- 游戏步骤：
- 将一条边删除
- 选择一条边E及由E连接的2个顶点V1和V2
- 用一个新的顶点取代边E以及由E连接着的2个顶点V1和V2。将由顶点V1和V2的整数值通过边E上的运算得到的结果赋予新顶点
- 重复以上步骤，直到所有边都被删除，游戏结束。游戏的得分就是所剩顶点上的整数值
我们的问题是： 根据给定的多边形，计算最高分和最底分
问题分析
- 最优子结构性质
- 这里呢，它是满足最优子结构性质的，我们不做过多的解释，直接看它的求解方法
- 递归求解
首先，我们在p(i, j)在op[i+s]处断开， 最大值记为maxf(i,j,s)，最小值记为minf(i,j,s)
因为只有加法和乘法，加法比较简单。对于乘法，可能存在负数，所以我们要对所有可能的结果做讨论，为了计算方便，我们可以定义如下表示
a=m[i,i+s,0] b=m[i,i+s,1]
c=m[i+s,j-s,0] d=m[i+s,j-s,1]当op[i+s]=‘+’
m[i,j,0]=a+c m[i,j,1]=b+d当op[i+s]=‘*’
m[i,j,0]=min{ac,ad,bc,bd}
m[i,j,1]=max{ac,ad,bc,bd}因此可有如下公式
至于s的断开位置，可以取 1 到 j - 1， 如下
接下来，就是超超超超级详细的解题步骤
超详细解题步骤
这里，我们举了这样一个栗子
详细步骤如下
m(1,1,1) = 9
m(1,1,0) = 9
m(2,1,1) = -4
m(2,1,0) = -4
m(3,1,1) = 9
m(3,1,0) = 9
m(4,1,1) = 10
m(4,1,0) = 10
m(5,1,1) = -10
m(5,1,0) = -10
m(1,2) = m(1,1)op(2)m(2,1)
op(2) = “+”
m(1,2,1) = m(1,1,1) + m(2,1,1) = 5
m(1,2,0) = m(1,1,0) + m(2,1,0) = 5
m(2,2) = m(2,1)op(3)m(3,1)
op(3) = “+”
m(2,2,1) = m(2,1,1) + m(3,1,1) = 5
m(2,2,0) = m(2,1,0) + m(3,1,0) = 5
m(3,2) = m(3,1)op(4)m(4,1)
op(4) = “+”
m(3,2,1) = m(3,1,1) + m(4,1,1) = 19
m(3,2,0) = m(3,1,0) + m(4,1,0) = 19
m(4,2) = m(4,1)op(5)m(5,1)
op(5) = “*”
m(4,2,1) = max{m(4,1,1)*m(5,1,1), m(4,1,1)*m(5,1,0), m(4,1,0)*m(5,1,1), m(4,1,0)*m(5,1,0)} = -100
m(4,2,0) = min{m(4,1,1)*m(5,1,1), m(4,1,1)*m(5,1,0), m(4,1,0)*m(5,1,1), m(4,1,0)*m(5,1,0)} = -100
m(5,2) = m(5,1)op(1)m(1,1)
m(5,2,1) = m(5,1,1) + m(1,1,1) = -1
m(5,2,0) = m(5,1,0) + m(1,1,0) = -1
m(1,3)={m(1,1)op(2)m(2,2)m(1,2)op(3)m(3,1)m(1,3) = \begin{cases} m(1,1)op(2)m(2,2)\\ m(1,2)op(3)m(3,1) \end{cases}m(1,3)={m(1,1)op(2)m(2,2)m(1,2)op(3)m(3,1)
op(2) = “+”
op(3) = “+”
m(1,3,1) = max{m(1,1,1)+m{2,2,1}, m(1,2,1)+m(3,1,1)} = 14 (m(1,1,1)+m(2,2,1))
m(1,3,0) = min{m(1,1,0)+m{2,2,0}, m(1,2,0)+m(3,1,0)} = 14 (m(1,1,0)+m(2,2,0))
m(2,3)={m(2,1)op(3)m(3,2)m(2,2)op(4)m(4,1)m(2,3) = \begin{cases} m(2,1)op(3)m(3,2)\\ m(2,2)op(4)m(4,1) \end{cases}m(2,3)={m(2,1)op(3)m(3,2)m(2,2)op(4)m(4,1)
op(3) = “+”
op(4) = “+”
m(2,3,1) = max{m(2,1,1)+m(3,2,1), m(2,2,1)+m(4,1,1)} = 15 (m(2,1,1)+m(3,2,1))
m(2,3,0) = min{m(2,1,0)+m(3,2,0), m(2,2,0)+m(4,1,0)} = 15 (m(2,1,0)+m(3,2,0))
m(3,3)={m(3,1)op(4)m(4,2)m(3,2)op(5)m(5,1)m(3,3) = \begin{cases} m(3,1)op(4)m(4,2)\\ m(3,2)op(5)m(5,1) \end{cases}m(3,3)={m(3,1)op(4)m(4,2)m(3,2)op(5)m(5,1)
op(4) = “+”
op(5) = “*”
m(3,3,1) = max{m(3,1,1)+m(4,2,1), max{m(3,2,1)*m(5,1,1), m(3,2,1)*m(5,1,0), m(3,2,0)*m(5,1,1), m(3,2,0)*m(5,1,0)}} = -91 (m(3,1,1)+m(4,2,1))
m(3,3,0) = min{m(3,1,0)+m(4,2,0), min{m(3,2,1)*m(5,1,1), m(3,2,1)*m(5,1,0), m(3,2,0)*m(5,1,1), m(3,2,0)*m(5,1,0)}} = -190 (m(3,2,1)*m(5,1,1))
m(4,3)={m(4,1)op(5)m(5,2)m(4,2)op(1)m(1,1)m(4,3) = \begin{cases} m(4,1)op(5)m(5,2)\\ m(4,2)op(1)m(1,1) \end{cases}m(4,3)={m(4,1)op(5)m(5,2)m(4,2)op(1)m(1,1)
op(5) = “*”
op(1) = “+”
m(4,3,1) = max{max{m(4,1,1)*m(5,2,1), m(4,1,1)*m(5,2,0), m(4,1,0)*m(5,2,1), m(4,1,0)*m(5,2,0)}, m(4,2,1)+m(1,1,1)} = -10 (m(4,1,1)*m(5,2,1))
m(4,3,0) = min{min{m(4,1,1)*m(5,2,1), m(4,1,1)*m(5,2,0), m(4,1,0)*m(5,2,1), m(4,1,0)*m(5,2,0)}, m(4,2,0)+m(1,1,0)} = -91 (m(4,2,0)+m(1,1,0))
m(5,3)={m(5,1)op(1)m(1,2)m(5,2)op(2)m(2,1)m(5,3) = \begin{cases} m(5,1)op(1)m(1,2)\\ m(5,2)op(2)m(2,1) \end{cases}m(5,3)={m(5,1)op(1)m(1,2)m(5,2)op(2)m(2,1)
op(1) = “+”
op(2) = “+”
m(5,3,1) = max{m(5,1,1)+m(1,2,1), m(5,2,1)+m(2,1,1)} = -5 (m(5,1,1)+m(1,2,1))
m(5,3,0) = min{m(5,1,0)+m(1,2,0), m(5,2,0)+m(2,1,0)} = -5 (m(5,1,0)+m(1,2,0))
m(1,4)={m(1,1)op(2)m(2,3)m(1,2)op(3)m(3,2)m(1,3)op(4)m(4,1)m(1,4) = \begin{cases} m(1,1)op(2)m(2,3)\\ m(1,2)op(3)m(3,2)\\ m(1,3)op(4)m(4,1) \end{cases}m(1,4)=⎩⎪⎨⎪⎧m(1,1)op(2)m(2,3)m(1,2)op(3)m(3,2)m(1,3)op(4)m(4,1)
op(2) = “+”
op(3) = “+”
op(4) = “+”
m(1,4,1) = max{m(1,1,1)+m(2,3,1), m(1,2,1)+m(3,2,1), m(1,3,1)+m(4,1,1)} = 24 (m(1,1,1)+m(1,2,1))
m(1,4,0) = min{m(1,1,0)+m(2,3,0), m(1,2,0)+m(3,2,0), m(1,3,0)+m(4,1,0)} = 24 (m(1,1,0)+m(2,3,0))
m(2,4)={m(2,1)op(3)m(3,3)m(2,2)op(4)m(4,2)m(2,3)op(5)m(5,1)m(2,4) = \begin{cases} m(2,1)op(3)m(3,3)\\ m(2,2)op(4)m(4,2)\\ m(2,3)op(5)m(5,1) \end{cases}m(2,4)=⎩⎪⎨⎪⎧m(2,1)op(3)m(3,3)m(2,2)op(4)m(4,2)m(2,3)op(5)m(5,1)
op(3) = “+”
op(4) = “+”
op(5) = “*”
m(2,4,1) = max{m(2,1,1)+m(3,3,1), m(2,2,1)+m(4,2,1), max{m(2,3,1)*m(5,1,1), m(2,3,1)*m(5,1,0), m(2,3,0)*m(5,1,1), m(2,3,0)*m(5,1,0)}} = -95 (m(2,1,1)+m(3,3,1))
m(2,4,0) = min{m(2,1,0)+m(3,3,0), m(2,2,0)+m(4,2,0), min{m(2,3,1)*m(5,1,1), m(2,3,1)*m(5,1,0), m(2,3,0)*m(5,1,1), m(2,3,0)*m(5,1,0)}} = -194 (m(2,1,0)+m(3,3,0))
m(3,4)={m(3,1)op(4)m(4,3)m(3,2)op(5)m(5,2)m(3,3)op(1)m(1,1)m(3,4) = \begin{cases} m(3,1)op(4)m(4,3)\\ m(3,2)op(5)m(5,2)\\ m(3,3)op(1)m(1,1) \end{cases}m(3,4)=⎩⎪⎨⎪⎧m(3,1)op(4)m(4,3)m(3,2)op(5)m(5,2)m(3,3)op(1)m(1,1)
op(4) = “+”
op(5) = “*”
op(1) = “+”
m(3,4,1) = max{m(3,1,1)+m(4,3,1), max{m(3,2,1)*m(5,2,1), m(3,2,1)*m(5,2,0), m(3,2,0)*m(5,2,1), m(3,2,0)*m(5,2,0)}, m(3,3,1)+m(1,1,1)} = -1 (m(3,1,1)+m(4,3,1))
m(3,4,0) = min{m(3,1,0)+m(4,3,0), min{m(3,2,1)*m(5,2,1), m(3,2,1)*m(5,2,0), m(3,2,0)*m(5,2,1), m(3,2,0)*m(5,2,0)}, m(3,3,0)+m(1,1,0)} = -181 (m(3,3,0)+m(1,1,0))
m(4,4)={m(4,1)op(5)m(5,3)m(4,2)op(1)m(1,2)m(4,3)op(2)m(2,1)m(4,4) = \begin{cases} m(4,1)op(5)m(5,3)\\ m(4,2)op(1)m(1,2)\\ m(4,3)op(2)m(2,1) \end{cases}m(4,4)=⎩⎪⎨⎪⎧m(4,1)op(5)m(5,3)m(4,2)op(1)m(1,2)m(4,3)op(2)m(2,1)
op(5) = “*”
op(1) = “+”
op(2) = “+”
m(4,4,1) = max{max{m(4,1,1)*m(5,3,1), m(4,1,1)*m(5,3,0), m(4,1,0)*m(5,3,1), m(4,1,0)*m(5,3,0)}, m(4,2,1)+m(1,2,1), m(4,3,1)+m(2,1,1)} = -14 (m(4,3,1)+m(2,1,1))
m(4,4,0) = min{min{m(4,1,1)*m(5,3,1), m(4,1,1)*m(5,3,0), m(4,1,0)*m(5,3,1), m(4,1,0)*m(5,3,0)}, m(4,2,0)+m(1,2,0), m(4,3,0)+m(2,1,0)} = -95 (m(4,2,0)+m(1,2,0))
m(5,4)={m(5,1)op(1)m(1,3)m(5,2)op(2)m(2,2)m(5,3)op(3)m(3,1)m(5,4) = \begin{cases} m(5,1)op(1)m(1,3)\\ m(5,2)op(2)m(2,2)\\ m(5,3)op(3)m(3,1) \end{cases}m(5,4)=⎩⎪⎨⎪⎧m(5,1)op(1)m(1,3)m(5,2)op(2)m(2,2)m(5,3)op(3)m(3,1)
op(1) = “+”
op(2) = “+”
op(3) = “+”
m(5,4,1) = max{m(5,1,1)+m(1,3,1), m(5,2,1)+m(2,2,1), m(5,3,1)+m(3,1,1)} = 4 (m(5,1,1)+m(1,3,1))
m(5,4,0) = min{m(5,1,0)+m(1,3,0), m(5,2,0)+m(2,2,0), m(5,3,0)+m(3,1,0)} = 4 (m(5,1,0)+m(1,3,0))
m(1,5)={m(1,1)op(2)m(2,4)m(1,2)op(3)m(3,3)m(1,3)op(4)m(4,2)m(1,4)op(5)m(5,1)m(1,5) = \begin{cases} m(1,1)op(2)m(2,4)\\ m(1,2)op(3)m(3,3)\\ m(1,3)op(4)m(4,2)\\ m(1,4)op(5)m(5,1) \end{cases}m(1,5)=⎩⎪⎪⎪⎨⎪⎪⎪⎧m(1,1)op(2)m(2,4)m(1,2)op(3)m(3,3)m(1,3)op(4)m(4,2)m(1,4)op(5)m(5,1)
op(2) = “+”
op(3) = “+”
op(4) = “+”
op(5) = “*”
m(1,5,1) = max{m(1,1,1)+m(2,4,1), m(1,2,1)+m(3,3,1), m(1,3,1)+m(4,2,1), max{m(1,4,1)*m(5,1,1), m(1,4,1)*m(5,1,0), m(1,4,0)*m(5,1,1), m(1,4,0)*m(5,1,0)}} = -86 (m(1,1,1)+m(2,4,1))
m(1,5,0) = min{m(1,1,0)+m(2,4,0), m(1,2,0)+m(3,3,0), m(1,3,0)+m(4,2,0), min{m(1,4,1)*m(5,1,1), m(1,4,1)*m(5,1,0), m(1,4,0)*m(5,1,1), m(1,4,0)*m(5,1,0)}} = -240 (m(1,4,1)*m(5,1,1))
m(2,5)={m(2,1)op(3)m(3,4)m(2,2)op(4)m(4,3)m(2,3)op(5)m(5,2)m(2,4)op(1)m(1,1)m(2,5) = \begin{cases} m(2,1)op(3)m(3,4)\\ m(2,2)op(4)m(4,3)\\ m(2,3)op(5)m(5,2)\\ m(2,4)op(1)m(1,1) \end{cases}m(2,5)=⎩⎪⎪⎪⎨⎪⎪⎪⎧m(2,1)op(3)m(3,4)m(2,2)op(4)m(4,3)m(2,3)op(5)m(5,2)m(2,4)op(1)m(1,1)
op(3) = “+”
op(4) = “+”
op(5) = “*”
op(1) = “+”
m(2,5,1) = max{m(2,1,1)+m(3,4,1), m(2,2,1)+m(4,3,1), max{m(2,3,1)*m(5,2,1), m(2,3,1)*m(5,2,0), m(2,3,0)*m(5,2,1), m(2,3,0)*m(5,2,0)}, m(2,4,1)+m(1,1,1)} = -5 (m(2,1,1)+m(3,4,1))
m(2,5,0) = min{m(2,1,0)+m(3,4,0), m(2,2,0)+m(4,3,0), min{m(2,3,1)*m(5,2,1), m(2,3,1)*m(5,2,0), m(2,3,0)*m(5,2,1), m(2,3,0)*m(5,2,0)}, m(2,4,0)+m(1,1,0)} = -185 (m(2,1,0)+m(3,4,0))
m(3,5)={m(3,1)op(4)m(4,4)m(3,2)op(5)m(5,3)m(3,3)op(1)m(1,2)m(3,4)op(2)m(2,1)m(3,5) = \begin{cases} m(3,1)op(4)m(4,4)\\ m(3,2)op(5)m(5,3)\\ m(3,3)op(1)m(1,2)\\ m(3,4)op(2)m(2,1) \end{cases}m(3,5)=⎩⎪⎪⎪⎨⎪⎪⎪⎧m(3,1)op(4)m(4,4)m(3,2)op(5)m(5,3)m(3,3)op(1)m(1,2)m(3,4)op(2)m(2,1)
op(4) = “+”
op(5) = “*”
op(1) = “+”
op(2) = “+”
m(3,5,1) = max{m(3,1,1)+m(4,4,1), max{m(3,2,1)*m(5,3,1), m(3,2,1)*m(5,3,0), m(3,2,0)*m(5,3,1), m(3,2,0)*m(5,3,0)}, m(3,3,1)+m(1,2,1), m(3,4,1)+m(2,1,1)} = -5 (m(3,1,1)+m(4,4,1))
m(3,5,0) = min{m(3,1,0)+m(4,4,0), min{m(3,2,1)*m(5,3,1), m(3,2,1)*m(5,3,0), m(3,2,0)*m(5,3,1), m(3,2,0)*m(5,3,0)}, m(3,3,0)+m(1,2,0), m(3,4,0)+m(2,1,0)} = -185 (m(3,3,0)+m(1,2,0))
m(4,5)={m(4,1)op(5)m(5,4)m(4,2)op(1)m(1,3)m(4,3)op(2)m(2,2)m(4,4)op(3)m(3,1)m(4,5) = \begin{cases} m(4,1)op(5)m(5,4)\\ m(4,2)op(1)m(1,3)\\ m(4,3)op(2)m(2,2)\\ m(4,4)op(3)m(3,1) \end{cases}m(4,5)=⎩⎪⎪⎪⎨⎪⎪⎪⎧m(4,1)op(5)m(5,4)m(4,2)op(1)m(1,3)m(4,3)op(2)m(2,2)m(4,4)op(3)m(3,1)
op(5) = “*”
op(1) = “+”
op(2) = “+”
op(3) = “+”
m(4,5,1) = max{max{m(4,1,1)*m(5,4,1), m(4,1,1)*m(5,4,0), m(4,1,0)*m(5,4,1), m(4,1,0)*m(5,4,0)}, m(4,2,1)+m(1,3,1), m(4,3,1)+m(2,2,1), m(4,4,1)+m(3,1,1)} = 40 (m(4,1,1)*m(5,4,1))
m(4,5,0) = min{min{m(4,1,1)*m(5,4,1), m(4,1,1)*m(5,4,0), m(4,1,0)*m(5,4,1), m(4,1,0)*m(5,4,0)}, m(4,2,0)+m(1,3,0), m(4,3,0)+m(2,2,0), m(4,4,0)+m(3,1,0)} = -86 (m(4,2,0)+m(1,3,0))
m(5,5)={m(5,1)op(1)m(1,4)m(5,2)op(2)m(2,3)m(5,3)op(3)m(3,2)m(5,4)op(4)m(4,1)m(5,5) = \begin{cases} m(5,1)op(1)m(1,4)\\ m(5,2)op(2)m(2,3)\\ m(5,3)op(3)m(3,2)\\ m(5,4)op(4)m(4,1) \end{cases}m(5,5)=⎩⎪⎪⎪⎨⎪⎪⎪⎧m(5,1)op(1)m(1,4)m(5,2)op(2)m(2,3)m(5,3)op(3)m(3,2)m(5,4)op(4)m(4,1)
op(1) = “+”
op(2) = “+”
op(3) = “+”
op(4) = “+”
m(5,5,1) = max{m(5,1,1)+m(1,4,1), m(5,2,2)+m(2,3,1), m(5,3,1)+m(3,2,1), m(5,4,1)+m(4,1,1)} = 14 (m(5,1,1)+m(1,4,1))
m(5,5,0) = min{m(5,1,0)+m(1,4,0), m(5,2,0)+m(2,3,0), m(5,3,0)+m(3,2,0), m(5,4,0)+m(4,1,0)} = 14 (m(5,1,0)+m(1,4,0))
至此，终于计算完毕，那么：
max = max{m(1,5,1), m(2,5,1), m(3,5,1), m(4,5,1), m(5,5,1)} = 40
max = min{m(1,5,0), m(2,5,0), m(3,5,0), m(4,5,0), m(5,5,0)} = -240
最后，是Java代码的实现过程
Java代码实现
package PolygonGame;public class PolygonGame {static int n; // 边和点个数static int minf, maxf; static int[] v; // 点集static char[] op; // 边集static int[][][] m; // 存放最终计算结果static int[] result; // 存放最大值和最小值public static void main(String[] args) {n = 4;// 我的案例是: + -7 + 4 * 2 * 5v = new int[] {Integer.MIN_VALUE, -7, 4, 2, 5};op = new char[] {' ', '+', '+', '*', '*'};m = new int[n + 1][n + 1][2];for (int i = 1; i <= n; i++) {// m[i][j][1] 表示起点为i，长度为j的最大值m[i][1][1] = v[i];// m[i][j][0] 表示起点为i，长度为j的最小值m[i][1][0] = v[i];}polyMax();System.out.println("最大值: " + result[0] + "\n最小值: " + result[1]);}/*** * @param i: 链的起点* @param s: 断开位置* @param j: 链的长度*/public static void minMax(int i, int s, int j) {int[] e = new int[n + 1];// 在op(i+s) 处进行分割int a = m[i][s][0], // 左半部分最小值b = m[i][s][1], // 左半部分最大值r = (i + s - 1) % n + 1, // 取余，防止溢出c = m[r][j - s][0], // 右半部分最小值d = m[r][j - s][1]; // 右半部分最大值if(op[r] == '+') {// 对符号进行判断，加号和乘号的处理方式不同minf = a + c;maxf = b + d;} else {// 为乘号// maxf=max{ac, ad, bc, bd}// minf=min{ac, ad, bc, bd}e[1] = a * c;e[2] = a * d;e[3] = b * c;e[4] = b * d;minf = e[1];maxf = e[1];for (int k = 2; k < 5; k++) {// 查找最大最小值if (minf > e[k])minf = e[k];if (maxf < e[k])maxf = e[k];}}}public static void polyMax() {for(int j = 2; j <= n; j++) // 总长度遍历for(int i = 1; i <= n; i++)for(int s = 1; s < j; s++) { // 断开位置// 求m[i][j][1]和m[i][j][0]minMax(i, s, j);if(m[i][j][0] > minf)m[i][j][0] = minf;if(m[i][j][1] < maxf)m[i][j][1] = maxf;}result = new int[2];result[0] = m[1][n][1];result[1] = m[1][n][0];for(int i = 1; i <= n; i++) {if (result[0] < m[i][n][1]) result[0] = m[i][n][1];if (result[1] > m[i][n][0]) result[1] = m[i][n][0];}}
}最大值: 33
最小值: -35
好了，到此结束。欢迎大家关注我
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