Optimal Array Multiplication Sequence
Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu
Submit Status Practice UVA 348
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Description
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Given two arrays A and B, we can determine the array C = AB using the standard definition of matrix multiplication:
The number of columns in the A array must be the same as the number of rows in the B array. Notationally, let’s say that rows(A) and columns(A) are the number of rows and columns, respectively, in the A array. The number of individual multiplications required to compute the entire C array (which will have the same number of rows as A and the same number of columns as B) is then rows(A) columns(B) columns(A). For example, if A is a tex2html_wrap_inline67 array, and B is a tex2html_wrap_inline71 array, it will take tex2html_wrap_inline73 , or 3000 multiplications to compute the C array.
To perform multiplication of more than two arrays we have a choice of how to proceed. For example, if X, Y, and Z are arrays, then to compute XYZ we could either compute (XY) Z or X (YZ). Suppose X is a tex2html_wrap_inline103 array, Y is a tex2html_wrap_inline67 array, and Z is a tex2html_wrap_inline111 array. Let’s look at the number of multiplications required to compute the product using the two different sequences:
(XY) Z
tex2html_wrap_inline119 multiplications to determine the product (X Y), a tex2html_wrap_inline123 array.
Then tex2html_wrap_inline125 multiplications to determine the final result.
Total multiplications: 4500.
X (YZ)
tex2html_wrap_inline133 multiplications to determine the product (YZ), a tex2html_wrap_inline139 array.
Then tex2html_wrap_inline141 multiplications to determine the final result.
Total multiplications: 8750.
Clearly we’ll be able to compute (XY) Z using fewer individual multiplications.
Given the size of each array in a sequence of arrays to be multiplied, you are to determine an optimal computational sequence. Optimality, for this problem, is relative to the number of individual multiplications required.
Input
For each array in the multiple sequences of arrays to be multiplied you will be given only the dimensions of the array. Each sequence will consist of an integer N which indicates the number of arrays to be multiplied, and then N pairs of integers, each pair giving the number of rows and columns in an array; the order in which the dimensions are given is the same as the order in which the arrays are to be multiplied. A value of zero for N indicates the end of the input. N will be no larger than 10.
Output
Assume the arrays are named tex2html_wrap_inline157 . Your output for each input case is to be a line containing a parenthesized expression clearly indicating the order in which the arrays are to be multiplied. Prefix the output for each case with the case number (they are sequentially numbered, starting with 1). Your output should strongly resemble that shown in the samples shown below. If, by chance, there are multiple correct sequences, any of these will be accepted as a valid answer.
Sample Input
3
1 5
5 20
20 1
3
5 10
10 20
20 35
6
30 35
35 15
15 5
5 10
10 20
20 25
0
Sample Output
Case 1: (A1 x (A2 x A3))
Case 2: ((A1 x A2) x A3)
Case 3: ((A1 x (A2 x A3)) x ((A4 x A5) x A6))
//最优矩阵链乘，路径输出
//结合刘汝佳书上所讲，收获是路径的输出
//dp[i][j]=max{dp[i][k]+dp[k+1][j]+A(i-1)*A(k)*A(j)}
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=15;
const long long inf=9999999999;
int n;
int dp[maxn][maxn];
int p[maxn][maxn];
int left[maxn],right[maxn];
int cases;
void print(int x,int y){//递归回去，进行路径的输出
if(x==y){
printf("A%d",x+1);return ;
}
printf("(");
print(x,p[x][y]);//i-->k
printf(" x ");
print(p[x][y]+1,y);//k+1--->j
printf(")");
}
void solve(){
for(int i=0;i<n;i++){
for(int j=i;j<n;j++){
if(i==j)dp[i][j]=0;
else dp[i][j]=inf;
}
}
memset(p,0,sizeof(p));
for(int d=1;d<n;d++){
for(int i=0;i+d<n;i++){
int j=i+d;
for(int k=i;k<j;k++){
int tmp=dp[i][k]+dp[k+1][j]+left[i]*right[k]*right[j];
// printf("tmp %d\n",tmp);
if(dp[i][j]>tmp){
dp[i][j]=tmp;
p[i][j]=k;
}
}
}
}
printf("Case %d: ", cases++);
print(0,n-1);
puts("");
}
int main(){
cases=1;
while(~scanf("%d",&n)&&n){
for(int i=0;i<n;i++){
scanf("%d%d",&left[i],&right[i]);
}
solve();
}
return 0;
}版权声明：本文为博主原创文章，未经博主允许不得转载。