POJ 1651 (最优矩阵链乘)

今天药忘吃喽~ 2024-02-17 19:33 50阅读 0赞

Multiplication Puzzle

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;9575</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;5954</td> 
  </tr> 
 </tbody> 
</table>

Description

The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one card out of the row and scores the number of points equal to the product of the number on the card taken and the numbers on the cards on the left and on the right of it. It is not allowed to take out the first and the last card in the row. After the final move, only two cards are left in the row.  
  
The goal is to take cards in such order as to minimize the total number of scored points.  
  
For example, if cards in the row contain numbers 10 1 50 20 5, player might take a card with 1, then 20 and 50, scoring  
10\*1\*50 + 50\*20\*5 + 10\*50\*5 = 500+5000+2500 = 8000  
If he would take the cards in the opposite order, i.e. 50, then 20, then 1, the score would be  
1\*50\*20 + 1\*20\*5 + 10\*1\*5 = 1000+100+50 = 1150.

Input

The first line of the input contains the number of cards N (3 <= N <= 100). The second line contains N integers in the range from 1 to 100, separated by spaces.

Output

Output must contain a single integer - the minimal score.

Sample Input

6
    10 1 50 50 20 5

Sample Output

3650

分析：最优链乘矩阵的变形（没有明确说明是矩阵，输入有点差别，但做法一样）

状态方程：dp(i,j)=min(dp(i,k)+dp(k+1,j)+v(i)\*v(k+1)\*v(j+1)); 表示从i到j的最低消耗（i,j表示第几个矩阵，矩阵的总个数t=n-1;）最后答案就是dp(1,t);递推求解

AC代码：

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    #define INF 2147483647
    const int maxn=100+10;
    int dp[maxn][maxn];
    int v[maxn];
    int n;
    
    int main(){
    	while(scanf("%d",&n)==1){
    		for(int i=1;i<=n;i++)
    		 {
    		 scanf("%d",&v[i]);
    		 dp[i][i]=0;
    		}
    		
    		int t=n-1;
    		for(int len=2;len<=t;len++)  //矩阵的长度 
    		 for(int i=1;i<=t-len+1;i++)  //起始位置 
    		  {
    		  	int j=i+len-1;  //终点位置 
    		  	dp[i][j]=INF;
    		    for(int k=i;k<j;k++)
    		     dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]+v[i]*v[k+1]*v[j+1]);  
    		}
    		
    		printf("%d\n",dp[1][t]);
    	}
    	return 0;
    }