Poj 1651 Multiplication Puzzle（区间dp）

Multiplication Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10010		Accepted: 6188

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

题解：

区间dp。。。设dp[l][r]表示区间[l,r]的最优解，则状态转移如下：
1、当r-l=2时，也即只有三个数时，显然dp[l][r] = a[l]*a[l+1]*a[r]；
2、当r-l>2时，对区间的最后一个被拿走的数进行枚举，则dp[l][r] = min(dp[l][r], dp[l][i]+dp[i][r]+a[l]*a[i]*a[r])，其中l<i<r。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<deque>
#include<algorithm>
#include<string>
#include<stack>
#include<cmath>
using namespace std;
int ans,n;
int a[105];
int dp[105][105];
const int inf=0x3f3f3f3f;
int  dfs(int l,int r)
{
if(r-l<2) return 0;
if(r-l==2) return dp[l][r]=a[l]*a[l+1]*a[r]; //起始值
if (dp[l][r]!=inf) return dp[l][r];
for(int i=l+1;i<r;i++)
dp[l][r]=min(dp[l][r],dfs(l,i)+dfs(i,r)+a[l]*a[i]*a[r]);
return dp[l][r];
}
int main()
{
while(~scanf("%d",&n))
{
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
ans=0;
memset(dp,inf,sizeof(dp));
printf("%d\n",dfs(1,n));
}
return 0;
}