CSU-ACM2017暑假训练9-区间DP E - Multiplication Puzzle

E - Multiplication Puzzle

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

本题较易，主要解释状态转移方程：

dp[s][e]=min{dp[s][k]+sample[s]∗sample[k]∗sample[e]+dp[k][e]}其中，dp[s][e]表示位置s、e为首尾的区间所能对应的最低得分sample[]表示位置为k的牌对应的数值 由题可知，对任意一个区间而言，最左和最右的牌是永远在桌面上的。我们令位置 k 为区间 [s,e] 中最后一张牌的位置，那么显然，对于这个区间来说，取出位置 k 上的牌后，区间的最终分数就是状态转移方程中右侧最小值函数内的部分。为此，我们只需对每一个区间枚举出它的最小得分，区间长度逐渐加大，就可以最终得到区间 [1,n] 所对应的最低得分了。

#include <iostream> #include <cstdio> #include <algorithm> #include <vector> #include <cstring> #include <queue> using namespace std; const int INF = 0x7fffffff; int dp[104][104], sample[105]; int main(){ #ifdef TEST freopen("test.txt", "r", stdin); #endif // TEST int n; while(cin >> n){ memset(dp, 0, sizeof(dp)); for(int i = 0; i < n; i++){ cin >> sample[i]; } for(int len = 3; len <= n; len++){//区间长度 for(int s = 0, e = len-1; e < n; s++, e++){ dp[s][e] = INF; for(int k = s+1; k < e; k++) dp[s][e] = min(dp[s][e], dp[s][k]+sample[s]*sample[k]*sample[e]+dp[k][e]); } } cout << dp[0][n-1] << endl; } return 0; }