矩阵链乘问题

问题描述：
给定n个矩阵的链<A1,A2,...,An>,矩阵Ai的规模为pi×pi+1(1≤i≤n)，求完全括号化的方案，使得计算乘积A1,A2,...,An所需标量乘法的次数最少。
递推式：

m[i,j]={0,min{m[i,k]+m[k+1,j]+pipk+1pj+1},if i=j if i<j 

#include <iostream>#include <vector>#include <climits>using namespace std;const int N=6;const int msize[6][2]={{30,35},{35,15},{15,5},{5,10},{10,20},{20,25}};const int p[7]={30,35,15,5,10,20,25};//the size of matrix: eg. matrix[0]=30*35; matrix[i]=p[i]*p[i+1]int m[N][N];//m[i][j]=m[i][k]+m[k+1][j]+p[i]*p[k+1]*p[j+1]vector< vector<int> > s(N);//store the split positionvoid matrix_chain_order(const int p[],int n){     n--;//the number of the matrixes  int i,j,l,k;  for(i=0;i<n;i++)  m[i][i]=0;  for(l=2;l<=n;l++){//l is the chain length, the maximum is n-1    for(i=0;i<=n-l;i++)//i is the start matrix of the chain,the last start matrix of the l length chain is n-l-1    {      j=i+l-1;//if the start of the chain is i,then the end of the chain is j, because j-i+1=l.      m[i][j]=INT_MAX;        for(k=i;k<j;k++)      {        int q=m[i][k]+m[k+1][j]+p[i]*p[k+1]*p[j+1];        if(q<m[i][j])        {          m[i][j]=q;          s[i][j]=k;        //  cout<<k<<endl;        }      }                  }                 }   }int look_up_chain(vector< vector<int> > &m,vector< vector<int> > &s,const int p[],int i,int j){    int k,q;    if(m[i][j]<INT_MAX)    return m[i][j];    if(i==j)    {        m[i][j]=0;    }else{        for(k=i;k<j;k++)        {            q=look_up_chain(m,s,p,i,k)+look_up_chain(m,s,p,k+1,j)+p[i]*p[k+1]*p[j+1];            if(q<m[i][j])            {                m[i][j]=q;                s[i][j]=k;            }        }    }    return m[i][j];}//带备忘机制的动态规划int memorized_matrix_chain(const int p[],int n){    n--;//the total matrixes in the chain    vector< vector<int> > memo(n);    for(int i=0;i<n;i++)    memo[i].resize(n);    int i,j;    for(i=0;i<n;i++)    {        for(j=i;j<n;j++)        memo[i][j]=INT_MAX;    }    return look_up_chain(memo,s,p,0,n-1);}void print_optimal_parens(vector< vector<int> > & s,int i, int j){     if(i==j)     cout<<"A_"<<i+1;     else{          cout<<"(";      print_optimal_parens(s,i,s[i][j]);      print_optimal_parens(s,s[i][j]+1,j);      cout<<")";     }}int main(){    for(int i=0;i<N;i++)    s[i].resize(N);   // matrix_chain_order(p,N+1);   int ans=memorized_matrix_chain(p,N+1);    cout<<"the answer: "<<endl;    cout<<ans<<endl;    print_optimal_parens(s,0,N-1);    return 0;}

用于实验的例子：

结果应为：((A1(A2A3))((A4A5)A6))
运行过程：

运行结果：