poj 1651 Multiplication Puzzle 区间dp

题目链接

给一串数， 拿出一个数i的代价是a[i-1]*a[i]*a[i+1]， 第一个数和最后一个数不能拿， 求拿走剩下的数的代价的最小值。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, a, n) for(int i = a; i<n; i++)
#define ull unsigned long long
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1e7;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
int dp[105][105], a[105];
int dfs(int l, int r) {
    if(~dp[l][r])
        return dp[l][r];
    dp[l][r] = inf;
    if(r-l==1)
        return dp[l][r] = 0;
    if(r-l == 2)
        return dp[l][r] = a[l]*a[l+1]*a[r];
    for(int i = l+1; i<r; i++) {
        dp[l][r] = min(dp[l][r], dfs(l, i)+dfs(i, r)+a[i]*a[l]*a[r]);
    }
    return dp[l][r];
}
int main()
{
    int n;
    while(cin>>n) {
        mem1(dp);
        for(int i = 1; i<=n; i++)
            scanf("%d", &a[i]);
        int ans = dfs(1, n);
        cout<<ans<<endl;
    }
    return 0;
}