标签：compute can NPU nes cti tag mat limit body
Source
https://onlinejudge.u-aizu.ac.jp/courses/lesson/1/ALDS1/all/ALDS1_10_B
Description
The goal of the matrix-chain multiplication problem is to find the most efficient way to multiply given nn matrices M1,M2,M3,...,MnM1,M2,M3,...,Mn.
Write a program which reads dimensions of MiMi, and finds the minimum number of scalar multiplications to compute the maxrix-chain multiplication M1M2...MnM1M2...Mn.
In the first line, an integer nn is given. In the following nn lines, the dimension of matrix MiMi (i=1...ni=1...n) is given by two integers rr and cc which respectively represents the number of rows and columns of MiMi.
Print the minimum number of scalar multiplication in a line.
6 30 35 35 15 15 5 5 10 10 20 20 25
15125
Code
1 #include<stdio.h> 2 #include<string.h> 3 #define M 103 4 #define Min 1000000000 5 void Matrix_Chain_Order(int p[],int n); 6 int main() 7 { 8 int n; 9 while(scanf("%d",&n) == 1){ 10 int p[M]; 11 memset(p,‘0‘,M); 12 int i; 13 for(i = 0;i < n;i++) 14 scanf("%d%d",&p[i],&p[i+1]); 15 Matrix_Chain_Order(p,n); 16 } 17 18 return 0; 19 } 20 21 void Matrix_Chain_Order(int p[],int n) 22 { 23 int m[n+1][n+1],s[n+1][n+1]; 24 //边界条件 25 int i,j; 26 for(i = 1;i <= n;i++) 27 m[i][i] = 0; 28 //计算每个子问题m[i,j]的解 29 int l; 30 for(l = 2;l <= n;l++){ 31 for(i = 1;i <= n-l+1;i++){ 32 j = i + l - 1; 33 m[i][j] = Min; 34 int k; 35 for(k = i;k <= j-1;k++){ 36 int val; 37 val = m[i][k] + m[k+1][j] + p[i-1]*p[k]*p[j]; 38 if(m[i][j] > val){ 39 m[i][j] = val; 40 s[i][j] = k; 41 } 42 } 43 } 44 } 45 printf("%d\n",m[1][n]); 46 }
标签：compute can NPU nes cti tag mat limit body
原文地址：https://www.cnblogs.com/zjsyzmx0527/p/9893378.html
