Poj-1651-multiplication puzzle

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Multiplication puzzle

Time limit:1000 ms		Memory limit:65536 K
Total submissions:6162		Accepted:3758

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 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 wocould take the cards in the opposite order, I. e. 50, then 20, then 1, the score wocould 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

Question: Give a string of numbers, and then give you an operation to select the numbers of the two non-edge numbers, then we get the score that is the product of this number and the two left and right numbers, and then the selected number disappears, asking when the total score is the maximum when there are only two left.
This is the interval DP. We can enumerate a certain number and then find the maximum score if the number disappears.
State transition equation: DP [I] [J] [k] = max (DP [I] [J] [K], (I + 1 ~ Maximum Value of J-1) + (J + 1 ~ K-1 value) + I * j * K)
It is convenient to use DFS to calculate the value, and the recursive depth is not long, so there is no need to worry about stack explosion or time issues.
Another reason for implementing recursive methods is that I usually prefer to use recursive methods when writing DP, but it is not enough to master only one implementation method, after all, different implementations may have different effects in different situations.

Code:

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #define MAX 102 6 #define LL long long 7 using namespace std; 8  9 const int limit = 100000002;10 int p[MAX];11 int dp[MAX][MAX][MAX];12 int n;13 14 int dfs(int a,int b,int c){15     int minn=limit;16     if(dp[a][b][c]!=-1) return dp[a][b][c];17     dp[a][b][c]=p[a]*p[b]*p[c];18     for(int i=a+1;i<b;i++){19         dfs(a,i,b);20         minn = min(minn,dp[a][i][b]);21     }22     if(a+1<b) dp[a][b][c]+=minn;23     minn = limit;24     for(int i=b+1;i<c;i++){25         dfs(b,i,c);26         minn = min(minn,dp[b][i][c]);27     }28     if(b+1<c) dp[a][b][c]+=minn;29     return dp[a][b][c];30 }31 32 int main()33 {34     int sum;35     //freopen("data.txt","r",stdin);36     while(~scanf("%d",&n)){37         for(int i=1;i<=n;i++) scanf("%d",&p[i]);38         memset(dp,-1,sizeof(dp));39         sum=limit;40         for(int i=2;i<n;i++){41             dfs(1,i,n);42             sum = min(sum,dp[1][i][n]);43         }44         printf("%d\n",sum);45     }46     return 0;47 }

1651