【POJ 1651】【区间DP 矩阵链乘的变形】Multiplication Puzzle【一串数字，除了头尾不能动，每次取个数字，它与左右相邻数字的乘积为其价值，求价值和最小】

传送门：POJ 1651

描述：

Multiplication Puzzle

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8754 Accepted: 5479

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

610 1 50 50 20 5

Sample Output

3650

Source

Northeastern Europe 2001, Far-Eastern Subregion

题意：

一系列的数字，除了头尾不能动，每次取出一个数字，这个数字与左右相邻数字的乘积为其价值，最后将所有价值加起来，要求最小值

思路：

这些数字其实就是n-1个矩阵的行列的数展开的排列，按照矩阵链乘那样区间DP就好了

代码：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <queue>#include <vector>#include <map>#include <set>#include <stack>#include <sstream>#include <cmath>#include <cstdlib>#include<string>#include <ctime>#define mst(ss,b) memset(ss,b,sizeof(ss));using  namespace std;const int inf=0x3f3f3f3f;const int N=105;int n,a[N],dp[N][N];int  main(){  #ifndef ONLINE_JUDGE  freopen("in.txt","r",stdin);  #endif  while(~scanf("%d",&n)){  for(int i=0; i<n; i++)scanf("%d",&a[i]);  mst(dp, 0);  for(int l=1; l<n; l++)  for(int i=1,j=l+i; j<n; i++,j++){  int mn=inf;  for(int k=i; k<j; k++)  mn=min(mn, dp[i][k]+dp[k+1][j]+a[i-1]*a[k]*a[j]);  dp[i][j]=mn;  }  printf("%d\n", dp[1][n-1]);//第一个矩阵到最后一个矩阵相乘  }  return 0;}