poj-1651-Multiplication Puzzle--区间DP

poj-1651-Multiplication Puzzle--区间DP

 

Multiplication Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10734		Accepted: 6704

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

 

 

1651

Accepted

508K

0MS

G++

764B

2017-09-16 15:26:12

 

对于这种区间DP问题，主要是写对DP方程式。开始写成 dp[i][j] = dp[i][k-1] + num[k-1]*num[k]*num[k+1] + dp[k+1][j] . 明显是没有理解题意。

将该num[k] 给取出之后， num[k-1] 和 num[k+1] 成为相邻的，所以：

不应该是需要从后面逆推，即是 dp[i][j] = dp[i][k] + dp[k][j] + num[i]*num[j]*num[k] . 可以从三个数字，四个数字这样试下。

 

#include <cstdio> 
#include <cstring>

#define min(a, b) (a)>(b)?(b):(a) 
const int MAXN = 110; 

int n, num[MAXN]; 
long long dp[MAXN][MAXN]; 

long long Solve(int l, int r){
	if(dp[l][r] != 0x3f3f3f3f3f3f){
		return dp[l][r]; 
	}
	if(l + 1 >= r){
		dp[l][r] = 0; 
		return 0; 
	}
	for(int i=l+1; i<r; ++i){
		long long tmp = num[l]*num[i]*num[r]; 
		dp[l][r] = min(dp[l][r], tmp + Solve(l, i) + Solve(i, r)); 
	} 
	return dp[l][r]; 
}

int main(){
	freopen("in.txt", "r", stdin); 

	long long ans; 
	while(scanf("%d", &n) != EOF){
		for(int i=0; i<n; ++i){
			scanf("%d", &num[i]);  
		} 
		for(int i=0; i<n; ++i){
			for(int j=0; j<n; ++j){
				dp[i][j] = 0x3f3f3f3f3f3f; 
			}
		}
		// memset(dp, 0x3f3f3f3f, sizeof(dp));  
		ans = Solve(0, n-1); 
		printf("%lld\n", ans );
	} 
	return 0; 
}