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poj 1651 Multiplication Puzzle（区间dp 矩阵链乘法）

2015-06-17 10:14 274 查看

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Multiplication Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6965		Accepted: 4268

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

给出一些数除了第一个跟最后一个其他的数任意取出取出一个数时有一个代价

代价值为这个数与相邻两个数的乘积

这个题就是类似于矩阵乘法计算次数

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>

#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define MAXN 10010
#define MAXM 100010
#define INF 99999999
#define ll __int64
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)

using namespace std;

int Read()
{
    char c = getchar();
    while (c < '0' || c > '9') c = getchar();
    int x = 0;
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x;
}

void Print(int a)
{
     if(a>9)
         Print(a/10);
     putchar(a%10+'0');
}

int dp[110][110],a[110];

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

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