POJ 1651 Multiplication Puzzle

最新推荐文章于 2023-08-28 09:42:08 发布

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最新推荐文章于 2023-08-28 09:42:08 发布

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分类专栏：区间DP

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本文介绍了一道典型的动态规划题目，通过将大区间不断分割成子区间，利用状态转移方程来求解移除数列中指定位置的数时能得到的最小累积得分。

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A - Multiplication Puzzle

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1651

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

区间DP。

题目的意思是，给一数序列。头尾不能移除，其他的数里，每移除一个数，将它左右的数和它自己相乘，得到一个值。最后累加这个值，求最小值。

典型的动态规划题目。

方法是把大区间【1，n】分成子区间。用[i,j]表示子区间。

那么可以得到状态转移方程。

dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+a[i]*a[k]*a[j]);

k为[i,j]中子区间分隔，

i<k<j;

将[i,j]区间不断的分成子区间[i,k]和[k,j]求最小值。

最后dp[1][n]
即为整个区间的最小值。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define N 105
#define INF 0x3f3f3f3f
using namespace std;
int a[N];
int dp[N][N];
int main()
{
        int n;
        while(scanf("%d",&n)>0)
        {
                memset(dp,0,sizeof(dp));
                for(int i=1;i<=n;i++)
                        scanf("%d",&a[i]);
                for(int d=2;d<=n-1;d++)
                        for(int i=1;i+d<=n;i++)
                        {
                                int j=i+d;
                                dp[i][j]=INF;
                                for(int k=i+1;k<=j-1;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][n]);
        }
        return 0;
}