poj1651（区间dp）

Multiplication Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7255		Accepted: 4481

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

Source

Northeastern Europe 2001, Far-Eastern Subregion

分析：区间从小到大，每次取出一个点，就相当于取某个已有区间的端点，起花费为该点和新区间的两端点的积，然后在遍历一次，找出该区间的最优解。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
#define ll long long
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;

int n,a[105],dp[105][105];

int main ()
{
    scanf ("%d",&n);
    for (int i=1; i<=n; i++)
        scanf ("%d",&a[i]);
    for (int i=n-2; i>0; i--)
    {
        dp[i][i+2] = a[i]*a[i+1]*a[i+2];
        for (int j=i+3; j<=n; j++)
        {
            dp[i][j] = min(dp[i+1][j]+a[i]*a[i+1]*a[j], dp[i][j-1]+a[i]*a[j-1]*a[j]);
            for (int k=i+2; k<=j-2; k++)
                dp[i][j] = min(dp[i][j], dp[i][k]+dp[k][j]+a[i]*a[k]*a[j]);
        }
    }
    printf ("%d\n",dp[1]
);
    return 0;
}