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[105],DP[105][105],N;

int main()
{
    while(~scanf("%d", &N))
    {
        for(int i=0;i<=N;i++)
        {
            for(int j=0;j<=N;j++) DP[i][j] = INF;
        }
        for(int i=1;i<=N;i++)
        {
            scanf("%d", &a[i]);
        }
        for(int i=2;i<N;i++)
        {
            for(int j=i;j>=2;j--)
            {
                if(i == j) DP[j][i] = a[i] * a[i-1] * a[i+1];
                else
                {
                    int x = DP[j+1][i]+a[j]*a[j-1]*a[i+1];
                    int y = DP[j][i-1]+a[i]*a[j-1]*a[i+1];
                    DP[j][i] = MIN(x, y);
                    if(i-j>1)for(int k = j+1;k<i;k++)
                    {
                        x = DP[j][k-1] + DP[k+1][i] + a[k]*a[j-1]*a[i+1];
                        DP[j][i] = MIN(DP[j][i], x);
                    }
                }
            }
        }
        printf("%d
", DP[2][N-1]);
    }
    return 0;
}