POJ 1651 Multiplication Puzzle

一、题目

1、题目描述

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.

2、输入/输出描述

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.

3、原题链接

1651 -- Multiplication Puzzle (poj.org)

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二、解题报告

1、思路分析

区间DP板子题，我们采用记忆化搜索来计算状态，即dfs(l,r)为抽走第l张到第r张卡牌获取的最小总点数，用二维数组f[l][r]来剪枝保存

数组a[]来保存每张牌的点数，初始化为-1

如果l > r，那么返回0

如果f[l][r]已经访问，那么直接返回

如果未访问，那么有f[l][r] = min(f[l][r] , dfs(l, i - 1) + dfs(i + 1, r) + abs(a[i] * a[l - 1] * a[r + 1]))

2、复杂度

时间复杂度： O(n^3) 空间复杂度：O(n^2)

3、代码详解

#include 
#include 
#include 
using namespace std;
#define N 210
int f[N][N], t, n;
int a[N];
const int inf = 0x3f3f3f3f3f3f3f3f;
int dfs(int l, int r)
{
    if (l > r)
        return 0;
    if (f[l][r] != inf)
        return f[l][r];

    int &res = f[l][r];
    for (int i = l; i <= r; i++)
        res = min(res, dfs(l, i - 1) + dfs(i + 1, r) + abs(a[i] * a[l - 1] * a[r + 1]));
    return res;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    memset(f, 0x3f, sizeof(f));
    memset(a, -1, sizeof(a));

    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    cout << dfs(2, n - 1);
    return 0;
}