poj1651

Multiplication Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7252		Accepted: 4478

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

3650

Source

Northeastern Europe 2001, Far-Eastern Subregion

区间dp,挺简单的。

给你一组数字，第一个和最后一个数字不可以取出去，其它任意取出去，当你要取出一个数字时，它有一个代价，这个代价就是与它相邻的两个数的乘积，求除了首位两位数字，把其他数字都取出来，它们的代价之和的最小值........

#include <cstdio>
#include <iostream>
#include <sstream>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;
#define ll long long
#define _cle(m, a) memset(m, a, sizeof(m))
#define repu(i, a, b) for(int i = a; i < b; i++)
#define repd(i, a, b) for(int i = b; i >= a; i--)
#define sfi(n) scanf("%d", &n)
#define pfi(n) printf("%d
", n)
#define sfi2(n, m) scanf("%d%d", &n, &m)
#define pfi2(n, m) printf("%d %d
", n, m)
#define pfi3(a, b, c) printf("%d %d %d
", a, b, c)
#define MAXN 105
const int INF = 0x3f3f3f3f;
int a[MAXN];
int dp[MAXN][MAXN];
int main()
{
    int n;
    while(~sfi(n))
    {
        repu(i, 0, n) sfi(a[i]);
        _cle(dp, 0x3f);
        dp[0][0] = a[0];
        a[n + 1] = 1;
        repu(i, 1, n + 1) dp[i][i] = a[i - 1] * a[i] * a[i + 1];
        repu(i, 1, n) repu(j, 0, i) dp[i][j] = 0;
        for(int i = n - 1; i > 0; i--)
        repu(j, i + 1, n) repu(k, i, j + 1)
           dp[i][j] = min(dp[i][j], dp[i][k - 1] + a[k] * a[i - 1] * a[j + 1] + dp[k + 1][j]);
        pfi(dp[1][n - 2]);
    }
    return 0;
}