POJ1651-Multiplication Puzzle

Multiplication Puzzle

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8669 Accepted: 5422

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：有N张写有数字的卡片排成一行，按一定次序从中拿走N-2张（第1张和最后一张不能拿），每次只拿一张，取走一张卡片的同时，会得到一个分数，分值的计算方法是：要拿的卡片，和它左右两边的卡片，这三张卡片上数字的乘积。按不同的顺序取走N-2张卡片，得到的总分可能不相同，求出给定一组卡片按上述规则拿取的最小得分。

#include <iostream>#include <cstdio>#include <cstring>#include <stack>#include <queue>#include <cmath>#include <algorithm>using namespace std;const int INF=0x3f3f3f3f;int a[110],dp[110][110];int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=0; i<n; i++)            scanf("%d",&a[i]);        memset(dp,0,sizeof dp);        for(int k=3;k<=n;k++)        {            for(int i=0;i<n-2;i++)            {                int j=k+i-1;                dp[i][j]=INF;                for(int t=i+1;t<j;t++)                    dp[i][j]=min(dp[i][j],dp[i][t]+dp[t][j]+a[i]*a[t]*a[j]);            }        }        printf("%d\n",dp[0][n-1]);    }    return 0;}