POJ

先上题目：

Multiplication Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
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 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

　　题意：给出一串数字，然后给你一种操作，选择非边缘的那两个数字的数字，然后获得值为这个数字与其左右两个数一共三个数的乘积的得分，然后被选的那个数消失，问最终只剩下两个数的时候总得分什么时候最大。
　　这一题是区间DP,我们可以枚举某一个数，然后求如果最终是这个数消失的话能得到的最大得分是多少就可以了。
　　状态转移方程：dp[i][j][k]=max(dp[i][j][k],(i+1~j-1的最值) + (j+1~k-1的最值) +i*j*k)
　　这里使用dfs来求值比较方便，而且递归的深度不会很长，所以不用担心爆栈或者时间的问题。
　　这里使用递归的方法实现的另一个原因的因为我平时写DP比较习惯用递推的形式，但是只掌握一种实现方式还是不够，毕竟不同的情况用不同的实现方式会有不一样的效果。

上代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAX 102
#define LL long long
using namespace std;

const int limit = 100000002;
int p[MAX];
int dp[MAX][MAX][MAX];
int n;

int dfs(int a,int b,int c){
    int minn=limit;
    if(dp[a][b][c]!=-1) return dp[a][b][c];
    dp[a][b][c]=p[a]*p[b]*p[c];
    for(int i=a+1;i<b;i++){
        dfs(a,i,b);
        minn = min(minn,dp[a][i][b]);
    }
    if(a+1<b) dp[a][b][c]+=minn;
    minn = limit;
    for(int i=b+1;i<c;i++){
        dfs(b,i,c);
        minn = min(minn,dp[b][i][c]);
    }
    if(b+1<c) dp[a][b][c]+=minn;
    return dp[a][b][c];
}

int main()
{
    int sum;
    //freopen("data.txt","r",stdin);
    while(~scanf("%d",&n)){
        for(int i=1;i<=n;i++) scanf("%d",&p[i]);
        memset(dp,-1,sizeof(dp));
        sum=limit;
        for(int i=2;i<n;i++){
            dfs(1,i,n);
            sum = min(sum,dp[1][i][n]);
        }
        printf("%d
",sum);
    }
    return 0;
}