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Multiplication Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11466		Accepted: 7105

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 題目大意：有n張卡片在一排，拿取n-2次，不能拿首尾的卡片，每次拿卡片時你都會獲得卡片前一張數值卡片本身數值卡片後一張卡片數值三個數的乘積當做你的得分。現在你需要以一種順序取走卡片使得你得到的分值最小，最後輸出能得到的最小的分值。 解題思路： 此處引用博客:https://www.cnblogs.com/Silenceneo-xw/p/5941353.html 一道區間DP的題目。設dp[l][r]表示區間[l,r]的最優解，則狀態轉移如下： 1、當r-l=2時，也即只有三個數時，顯然dp[l][r] = num[l]*num[l+1]*num[r]； 2、當r-l>2時，對區間的最後一個被拿走的數進行枚舉，則dp[l][r] = min(dp[l][r], dp[l][i]+dp[i][r]+num[l]*num[i]*num[r])，其中l<i<r。 AC代碼:

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int INF=0x3f3f3f3f;
const int MAXN=100+10;
int num[MAXN],dp[MAXN][MAXN];

int solve(int i,int j)
{
    if(dp[i][j]!=INF)return dp[i][j];
    if(j==i+1)return dp[i][j]=0;
    for(int k=i+1;k<j;k++)
        dp[i][j]=min(dp[i][j],num[i]*num[k]*num[j]+solve(i,k)+solve(k,j));
    return dp[i][j];
}

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&num[i]);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
            dp[i][j]=INF;
        printf("%d\n",solve(1,n));
    }
    return 0;
}

Multiplication Puzzle POJ - 1651(區間DP)