POJ1651:Multiplication Puzzle（区间dp）

最新推荐文章于 2021-05-25 11:51:06 发布

junior19

最新推荐文章于 2021-05-25 11:51:06 发布

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分类专栏：区间dp

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本文介绍了一种通过特定顺序取卡片上的整数来最小化总得分的算法。玩家需从一排含有正整数的卡片中按最优顺序取走卡片，使得基于所取卡片及其相邻卡片数值计算出的总得分最低。文章提供了实现该算法的C语言代码示例。

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Multiplication Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9419		Accepted: 5850

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

# include <stdio.h>
# include <string.h>
int min(int a, int b)
{
    return a<b?a:b;
}
int main()
{
    int n, len, i, k, imin1, imin2, imin, a[102],dp[102][102];
    while(scanf("%d",&n) != EOF)
    {
    memset(dp,0 ,sizeof(dp));
    for(i=0; i<n; ++i)
        scanf("%d",&a[i]);
    for(i=1; i<n-1; ++i)
        dp[i][i] = a[i]*a[i-1]*a[i+1];//初始化
    for(len=1; len<n; ++len)//枚举区间
        for(i=1; i+len<n-1; ++i)
        {
            //枚举最后消去的点
            imin1 = dp[i+1][i+len]+a[i]*a[i-1]*a[i+len+1];//（为首点）
            imin2 = dp[i][i+len-1]+a[i+len]*a[i-1]*a[i+len+1];//（为尾点）
            imin = min(imin1, imin2);
            for(k=i+1; k<i+len; ++k)//（为中间点）
                imin = min(imin, dp[i][k-1]+dp[k+1][i+len]+a[k]*a[i-1]*a[i+len+1]);
            dp[i][i+len] = imin;
        }
        printf("%d\n",dp[1][n-2]);
    }
    return 0;
}