Matrix chain multiplication problem

A good example of dynamic programming is an algorithm that solves the problem of matrix-chain multiplication. We are given a sequence (chain) 〈A₁, A₂, ..., A_n〉 of n matrices to be multiplied, and we wish to compute the product A₁ A₂... A_n.

 

#include<iostream>
#include<fstream>
#include<sstream>
#include<string>

#define MAX 99999

using namespace std;

int s[100][100];//s[i][j]记录Ai..Aj中分割位置
void print(int i,int j)
...{
    if(i == j)
        cout << 'A' << i;
    else
    ...{
        cout <<'(';
        print(i,s[i][j]);
        print(s[i][j]+1,j);
        cout << ')';
    }
}


int main()
...{
    int p[100];//记录维度，从下标0开始
    int len=-1;//记录p已用位置
    int m[100][100];//m[i][j]表示矩阵Ai..Aj所做乘运算最少的数目


    string str;
    int tmp;
    ifstream cin("1.txt");
    while(getline(cin,str))//一次处理一个实例
    ...{
        istringstream stream(str);
        while(stream >> tmp)
        ...{
            p[++len] = tmp;
        }
        for(int i = 1; i <= len; i++)
            m[i][i] = 0;

        for( i = 0; i <= len; i++)
            cout << p[i] << "  ";
        cout <<len<<endl;
        int j,k;
        int q;

        for(int l = 2; l <= len; l++)//for1,l is the chain length
        ...{
            for(i = 1; i <= len-l+1; i++)//for2,
            ...{
                j = i + l - 1;
                m[i][j] = MAX;
                for(k = i; k <= j-1; k++)//for3
                ...{
                    q = m[i][k] + m[k+1][j] + p[i-1]*p[k]*p[j];
                    if(q < m[i][j])
                    ...{
                        m[i][j] = q;
                        s[i][j] = k;
                    }
                }//for3
            }//for2
        }//for1

        print( 1,len);
        cout << endl;
        len = -1;
    }//while

    return 0;
}