POJ 1651 Multiplication Puzzle（類似矩陣連乘 區間dp）

傳送門：http://poj.org/problem?id=1651

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Multiplication Puzzle

Time Limit: 1000MS	?	Memory Limit: 65536K
Total Submissions: 13109	?	Accepted: 8034

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

題目意思：

給你一串數字，頭尾不能動，每次取出一個數字，這個數字貢獻=該數字與左右相鄰數字的乘積，求一個最小值。

分析：

是一個區間dp問題，類似矩陣連乘的做法，也是需要在一個區間中選擇一個k值從而來達到你的某種要求，這里是要使得消去的值最小

概況一下題意就是：

初使ans=0,每次消去一個值,位置在pos（pos!=1 && pos !=n）
同時ans+=a[pos-1]*a[pos]*a[pos+1]，一直消元素直到最后剩余2個，求方案最小的ans是多少？

#include <iostream>
#include<algorithm>
#include <cstdio>
#include<cstring>
using namespace std;
#define max_v 105
#define INF 9999999
int a[max_v];
int dp[max_v][max_v];
int main()
{int n;while(~scanf("%d",&n)){for(int i=1;i<=n;i++){scanf("%d",&a[i]);}memset(dp,0,sizeof(dp));for(int r=2;r<n;r++){for(int i=2;i<=n-r+1;i++){int j=i+r-1;int t=INF;for(int k=i;k<=j-1;k++){t=min(t,dp[i][k]+dp[k+1][j]+a[i-1]*a[k]*a[j]);}dp[i][j]=t;}}printf("%d\n",dp[2][n]);}return 0;
}

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