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
Source
Northeastern Europe 2001, Far-Eastern Subregion
题目类型：区间DP
算法分析：使用dp[i][j]表示区间(i,j)之间的数被拿去所具有的最小乘积值，边界条件是dp[i][i+1] = 0, dp[i][i+2] = a[i] * a[i+1]*a[i+2]，转移方程为dp[i][j] = min (dp[i][j], dp[i][k] + dp[k][j] + a[i]*a[k]*a[j]) i < k < j，最后输出dp[1][n]即可
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#include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("aaa.txt", "r", stdin) #define CPPFF ifstream cin ("aaa.txt") #define i64 long long const int INF = 0x7FFFFFFF - 6; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 200 + 13; int a[maxn], dp[maxn][maxn]; int main() { // CPPFF; int n; while (cin >> n) { for (int i = 1; i <= n; i++) cin >> a[i]; for (int i = 1; i <= n - 1; i++) dp[i][i+1] = 0; for(int i = 1; i <= n - 2; i++) dp[i][i+2] = a[i] * a[i+1] * a[i+2]; for (int l = 2; l <= n; l++) { for (int i = 1; i + l <= n; i++) { int j = i + l; dp[i][j] = INF; for (int k = i + 1; k < j; k++) dp[i][j] = min (dp[i][j], dp[i][k] + dp[k][j] + a[i] * a[k] * a[j]); } } cout << dp[1][n] << endl; } return 0; } |
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