［POJ-1651］Multiplication Puzzle [区间DP 入门]

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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
10150 + 50205 + 10505 = 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
15020 + 1205 + 1015 = 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
区间DP初步
$dp[width][L]$ 表示: 对区间 $[L, R] (R - L + 1= width)$进行如题意的操作最小分数为多少
$dp[width][L] = min(dp[width][L], dp[mid - L][L] + dp[R - mid][mid + 1] + val[mid] * val[L - 1] * val[R + 1])$
代码

 
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <string>
#include <math.h>
#include <stack>
#include <vector>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define rep(i, l, r) for(int i = l; i < r; i++)
#define per(i, r, l) for(int i = r; i >= l; i--)
#define dbg(...) cerr<<"["<<#__VA_ARGS__":"<<(__VA_ARGS__)<<"]"<<"\n"


typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int>pii;

const int N = (int) 100 + 11;
const int M = (int) 1e6 + 11;
const int MOD = (int) 1e9 + 7;
const int INF = (int) 0x3f3f3f3f;
const ll INFF = (ll) 0x3f3f3f3f3f3f3f3f;
/*-----------------------------------------------------------*/

ll val[N], dp[N][N];
int main(){
	int n; scanf("%d" ,&n);
	rep(i, 0, n)  scanf("%lld", val + i + 1);
	rep(i, 2, n) dp[1][i] = val[i] * val[i - 1] * val[i + 1]; // 初始化
	
	for(int width = 2; width <= n - 2; width++){
		for(int L = 2; L + width - 1 <= n - 1; L++){
			int R = L + width - 1;
			dp[width][L] = INFF;
			for(int mid = L; mid <= R; mid++){// mid 为枚举[L, R]中最后一个取出
				dp[width][L] = min(dp[width][L], dp[mid - L][L] + dp[R - mid][mid + 1] + val[mid] * val[L - 1] * val[R + 1]);
			}
		}
	} 
	cout << dp[n - 2][2] <<"\n";
	return 0;
}