AYITACM2016 Province third week i-optimal Array multiplication Sequence (DP)

Matrix least multiplication

Test instructions

Give you 2 matrices A, B, we use the standard matrix multiplication definition c=ab as follows:

The number of columns in a array (column) must be equal to the number of column (row) in the B array to multiply this 2 array. If we use rows (a), columns (a), respectively, to represent the number of columns and bars in the A array, the number of multiplication required for the C array to be calculated is: Rows (a) *columns (B) *columns (a). For example: A array is a 10x20 matrix, b array is a 20x15 matrix, then to calculate the C array needs to do 10*15*20, that is, 3,000 times multiplication.

To calculate the multiplication of more than 2 matrices, you have to decide what order to use. For example: X, Y, z are matrices, there are 2 options to calculate XYZ: (XY) Z or X (YZ). Suppose X is an array of 5x10, Y is an array of 10x20, Z is an array of 20x35, and the number of multiplication required for that different sequence of operations is different:

(XY) Z

• 5*20*10 = 1000 times The multiplication is done (XY), and a 5x20 array is obtained.
• 5*35*20 = 3,500 times The final result is obtained.
• Total number of multiplication required: 1000+3500=4500.

X (YZ)

• 10*35*20 = 7,000 times The multiplication is completed (YZ), and a 10x35 array is obtained.
• 5*35*10 = 1750 times The final result is obtained.
• Total number of multiplication required: 7000+1750=8750.

Obviously, we can see that the calculation (XY) z uses fewer times of multiplication.

The question is: give you some matrices, you have to write a program to decide how to multiply the order, so that the number of times to use the multiplication will be the least.

Idea: the optimal matrix chain multiplication problem, the typical dynamic programming topic.

#include <stdio.h> #define M 1000#define inf 0x3f3f3f3fint dp[m][m],s[m][m],p[m],n;void print (int i,int j)//        Recursively outputs all results {if (i==j) printf ("a%d", I);//output is the number of data else {printf ("(");        Print (i,s[i][j]);//recursive Next data printf ("X");        Print (S[I][J]+1,J);    printf (")");    }}void work () {int i,j,k,l,t;            for (I=1; i<=n; i++) dp[i][i]=0;//first initialized to 0 for (l=2; l<=n; l++) for (i=1; i<=n-l+1; i++) {         J=I+L-1;            Determine the optimal solution dp[i][j]=inf between I and J;                for (k=i; k<=j-1; k++) {t=dp[i][k]+dp[k+1][j]+p[i-1]*p[k]*p[j];  if (T<dp[i][j]) {dp[i][j]=t; Maintain optimal solution s[i][j]=k; Record where parentheses are required}}} print (1,n); The optimal solution between the output I and J puts ("");}    int main () {int c=1,i; while (scanf ("%d", &n) &&n) {for (i=1; i<=n; i++) scanf ("%d%d", &p[i-1],&p[i]);//Outputinto the data printf ("Case%d:", C + +);    Work (); } return 0;}

AYITACM2016 Province third week i-optimal Array multiplication Sequence (DP)