POJ 1651 - Multiplication Puzzle （区间DP）

Multiplication Puzzle Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 12296   Accepted: 7508 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

题意：

去除一个数a[i]，代价为a[i-1]*a[i]*a[i+1]。

最左最右不能去除。

问你去除到只有2个端点后代价最小

看代码

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<string>

using namespace std;
#define ll long long int
const int inf  = 0x3f3f3f3f;
int dp[111][111];//dp[x][y]代表  x到y区间 全部合并只剩x和y两个数的最小花费。
int a[111];
int main()
{
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
	for(int i=1;i<=n;i++){
		for(int j=i;j<=n;j++){
			dp[i][j]=i+1==j?0:inf;
		}
	}
	for(int len=3;len<=n;len++){
		for(int i=1;i<=n-len+1;i++){
			int to=i+len-1;
			for(int k=i+1;k<to;k++){
				dp[i][to]=min(dp[i][to],dp[i][k]+a[i]*a[k]*a[to]+dp[k][to]);
			}
		}

	}
	cout<<dp[1][n]<<endl;

	return 0;
}