POJ1651Multiplication Puzzle（区间DP）

比较好做的区间DP

状态转移方程：DP[i][j] 表示区间[i,j]最小的乘积和。

DP[i][j] = MIN{DP[i][k-1]+DP[k+1][j] + a[k]*a[i-1]*a[j+1] | i<k<j};

最后结果就是DP[2][N-1]

 #include <map>
 #include <set>
 #include <stack>
 #include <queue>
 #include <cmath>
 #include <ctime>
 #include <vector>
 #include <cstdio>
 #include <cctype>
 #include <cstring>
 #include <cstdlib>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 #define eps 1e-15
 #define MAXN 105
 #define INF 1000000007
 #define MAX(a,b) (a > b ? a : b)
 #define MIN(a,b) (a < b ? a : b)
 #define mem(a) memset(a,0,sizeof(a))

 int a[],DP[][],N;

 int main()
 {
     while(~scanf("%d", &N))
     {
         for(int i=;i<=N;i++)
         {
             for(int j=;j<=N;j++) DP[i][j] = INF;
         }
         for(int i=;i<=N;i++)
         {
             scanf("%d", &a[i]);
         }
         for(int i=;i<N;i++)
         {
             for(int j=i;j>=;j--)
             {
                 if(i == j) DP[j][i] = a[i] * a[i-] * a[i+];
                 else
                 {
                     int x = DP[j+][i]+a[j]*a[j-]*a[i+];
                     int y = DP[j][i-]+a[i]*a[j-]*a[i+];
                     DP[j][i] = MIN(x, y);
                     if(i-j>)for(int k = j+;k<i;k++)
                     {
                         x = DP[j][k-] + DP[k+][i] + a[k]*a[j-]*a[i+];
                         DP[j][i] = MIN(DP[j][i], x);
                     }
                 }
             }
         }
         printf("%d\n", DP[][N-]);
     }
     return ;
 }