uva 3485 Optimal Array Multiplication Sequence
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题目:
I - Optimal Array Multiplication Sequence
Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu
Submit Status Practice UVA 348
Description
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) andcolumns(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 array, and B is a array, it will take , 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 array, Y is a array, and Z is a array. Let's look at the number of multiplications required to compute the product using the two different sequences:
(XY) Z
· multiplications to determine the product (X Y), a array.
· Then multiplications to determine the final result.
· Total multiplications: 4500.
X (YZ)
· multiplications to determine the product (YZ), a array.
· Then 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 . 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))
题目大意:
给n矩阵的长宽如la wa,当两个矩阵相乘时,费用为la*wb*wa,求这N个方程相乘的最小值.
题目思路:
1、因为答案必然是两个矩阵相乘,所以我们就可以用dp[i][j]来求dp[i][k]+dp[k][j]+fei[i][j] 最小的值
题目优化:
1、如果只用最长单调子序列求,可能会超时,所以可以二分答案
2、但如果题目时间要求短用二分搜索答案还是可能超时
3、所以采用反最长单调递增子序列( 最长单调递整子序列)
4、分析:假设长按照单调递减来求,简化为下图.如果前面的高于后面的,后面必然会在他的子序列里面,否则就ans++; 这样就转化为求最长单调递增子序列
程序:
#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<cmath>#include<algorithm>#include<cctype>#include <fstream>#include <limits>#include <vector>#include <list>#include <set>#include <map>#include <queue>#include <stack>#include <cassert>using namespace std;int putout(int ,int );int ans[20][20];int solve();int n,ci;int dp[20][20];struct node{ int l,r;}a[20];int main(){ while(~scanf("%d",&n)&&n) { for(int i=0;i<n;i++) scanf("%d%d",&a[i].l,&a[i].r); solve(); putout(0,n-1); }return 0;}int solve(){ memset(dp,0,sizeof(dp)); // for(int i=0;i<n-1;i++)dp[i][i+1]=a[i].l*a[i+1].r*a[i].r; for(int cha=1;cha<n;cha++) for(int i=0;i+cha<n;i++)//先求差值小的低值,在往上推 { int j=i+cha,st,ed; for(int k=i;k<j;k++) { int t=dp[i][k]+dp[k+1][j]+a[i].l*a[j].r*a[k].r; if(t<dp[i][j]||dp[i][j]==0) { dp[i][j]=t; ans[i][j]=k; } } //cout<<'!'<<i<<j<<dp[i][j]<<endl; }}int putout(int i,int j){//cout<<'!'<<i<<j<<endl;getchar(); if(i==0&&j==n-1)printf("Case %d: ",++ci); if(i==j) printf("A%d",i+1); else { printf("("); putout(i,ans[i][j]); printf(" x "); putout(ans[i][j]+1,j); printf(")"); } if(i==0&&j==n-1) printf("\n");}
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