xtu read problem training 4 B - Multiplication Puzzle

Multiplication Puzzle

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 1651
64-bit integer IO format: %lld      Java class name: Main

 

 

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

 

解题：dp[i][j]表示从i到j被划分后的最小值！为什么dp[i][j] = min(dp[i][j],dp[i][k]+dp[k][j]+d[i]*d[j]*d[k])ne

 

举个栗子 1 2 3 4 5

 

dp[1][5] = min(dp[1][5],dp[1][3]+dp[3][5]+d[1]*d[3]*d[5]) dp[i][j]表示i j段 剩有i j,像刚才的转移方程，dp[1][5]不是取了3以后 剩下了1 5 么

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;
int dp[110][110],d[110],n;
int main(){
    int i,j,k;
    while(~scanf("%d",&n)){
        for(i = 1; i <= n; i++)
            scanf("%d",d+i);
        memset(dp,0,sizeof(dp));
        for(k = 3; k <= n; k++){
            for(i = 1; i+k-1 <= n; i++){
                dp[i][i+k-1] = INF;
                for(j = i+1; j < i+k; j++)
                    dp[i][i+k-1] = min(dp[i][i+k-1],dp[i][j]+dp[j][i+k-1]+d[i]*d[j]*d[i+k-1]);
            }
        }
        cout<<dp[1][n]<<endl;
    }
    return 0;
}