xtu read problem training 4 B - Multiplication Puzzle

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Multiplication PuzzleTime Limit: 1000msMemory Limit: 65536KBThis 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. InputThe 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. OutputOutput must contain a single integer - the minimal score. Sample Input

610 1 50 50 20 5

Sample Output

3650

SourceNortheastern Europe 2001, Far-Eastern Subregion 解題：dp[i][j]表示從i到j被劃分後的最小值！為什麼dp[i][j] = min(dp[i][j],dp[i][k]+dp[k][j]+d[i]*d[j]*d[k])ne 舉個栗子 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,像剛才的轉移方程，dp[1][5]不是取了3以後 剩下了1 5 麼 

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib>10 #include <string>11 #include <set>12 #include <stack>13 #define LL long long14 #define INF 0x3f3f3f3f15 using namespace std;16 int dp[110][110],d[110],n;17 int main(){18     int i,j,k;19     while(~scanf("%d",&n)){20         for(i = 1; i <= n; i++)21             scanf("%d",d+i);22         memset(dp,0,sizeof(dp));23         for(k = 3; k <= n; k++){24             for(i = 1; i+k-1 <= n; i++){25                 dp[i][i+k-1] = INF;26                 for(j = i+1; j < i+k; j++)27                     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]);28             }29         }30         cout<<dp[1][n]<<endl;31     }32     return 0;33 }

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