[poj1651][区间DP]Multiplication Puzzle-优快云博客

本文介绍了一道经典的区间动态规划题目——乘法谜题。玩家需从一排含有正整数的卡片中按顺序取走卡片以使总得分最低。文章通过C++代码详细解释了解决方案。

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Multiplication Puzzle
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 11480 Accepted: 7117
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

简要题意：
首尾不能去，去掉val[x]会产生val[x]*val[x-1]*val[x+1]的代价，求最小代价。
显然是个区间dp了。设f[l][r]表示l和r不去，其他去完的最小代价。转移时枚举一下这个区间最后一个取得数是谁，转移即可

#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
using namespace std;
int n,m;
inline int read()
{
    char c;
    int res,flag=0;
    while((c=getchar())>'9'||c<'0') if(c=='-')flag=1;
    res=c-'0';
    while((c=getchar())>='0'&&c<='9') res=(res<<3)+(res<<1)+c-'0';
    return flag?-res:res;
}
const int N=110;
int f[N][N],a[N];
bool vis[N][N];
inline void solve(int l,int r)
{
    if(r-l+1<3||vis[l][r]) return;
    vis[l][r]=1;
    if(r-l+1==3)
    {
        f[l][r]=a[l]*a[l+1]*a[r];
        return;
    }
    f[l][r]=1e9;
    for(int k=l+1;k<r;++k)
    {
        solve(l,k);
        solve(k,r);
        f[l][r]=min(f[l][r],f[l][k]+f[k][r]+a[k]*a[l]*a[r]);
    }
}
int main()
{
//  freopen("1651.in","r",stdin);
//  freopen(".out","w",stdout);
    n=read();
    for(int i=1;i<=n;++i) a[i]=read();
    solve(1,n);
    printf("%d\n",f[1][n]);
}