package com.cjs.algorithm; public class DynamicPlan { /**
* 此方法用来求解矩阵连乘的最小数乘次数
*
* @param p
* 传入的要连乘的矩阵的维数信息的数组
* @return String型的矩阵的最小数层次数信息
*/
public static String matrixChain(int p[]) { int n = p.length - 1; //为p的实际最大下标
int m[][] = new int[n + 1][n + 1];
int s[][] = new int[n + 1][n + 1]; for (int i = 1; i <= n; i++) {
m[i][i] = 0;
}
for (int r = 2; r <= n; r++) // r为当前计算的链长(子问题规模)
{
for (int i = 1; i <= n - r + 1; i++)// n-r+1为最后一个r链的前边界
{
int j = i + r - 1;// 计算前边界为r,链长为r的链的后边界 m[i][j] = m[i + 1][j] + p[i - 1] * p[i] * p[j];// 将链ij划分为A(i) *( A[i+1:j] )
s[i][j] = i; for (int k = i + 1; k < j; k++) {
// 将链ij划分为( A[i:k] )* (A[k+1:j])
int t = m[i][k] + m[k + 1][j] + p[i - 1] * p[k] * p[j];
if (t < m[i][j]) {
m[i][j] = t;
s[i][j] = k;
}
}
}
} String answer = "";
answer = answer + "此矩阵连乘所需的最小次数为:" + m[1][n] + "\n";
matrixTraceBack(1, n, s); return answer; } private static void matrixTraceBack(int i, int j, int s[][]) { if (i == j) {
return;
}
matrixTraceBack(i, s[i][j], s);
matrixTraceBack(s[i][j] + 1, j, s); int x = s[i][j] + 1;
System.out.print("Multipy A" + i + "," + s[i][j]);
System.out.println(" and A" + x + "," + j); } }