poj3186 dp

/************************************************ *Author        :jibancanyang *Created Time  :二  4/ 5 09:33:33 2016 *题目类型:*最直观的思维是发现搜索很好写,但是仔细一想是可以转化为递推的.递推的时候要找 *矩阵中的转移路线然后确定递推顺序.本题也是一直二叉树来记忆化逐步优化为递推二维dp的典型例子*************************************************/#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <cstdlib>#include <ctime>#include <stack>using namespace std;typedef pair<int, int> pii;typedef long long ll;typedef unsigned long long ull;vector<int> vi;#define xx first#define yy second#define sa(n) scanf("%d", &(n))#define rep(i, a, n) for (int i = a; i < n; i++)#define vep(c) for(decltype((c).begin() ) it = (c).begin(); it != (c).end(); it++) const int mod = int(1e9) + 7, INF = 0x3fffffff, maxn = 2e3 + 12;int a[maxn];int dp[maxn][maxn];int main(void){#ifdef LOCAL    //freopen("in.txt", "r", stdin);    //freopen("out.txt", "w", stdout);#endif    int n;    while (~sa(n)) {        rep (i, 1, n + 1) sa(a[i]), dp[i][i] = a[i] * n;        for (int i = n - 1; i > 0; i--) {            for (int j = i + 1; j <= n; j++) {                int t = n - j + i;                 dp[i][j] = max(dp[i][j - 1] + t * a[j], dp[i + 1][j] + t * a[i]);            }        }        cout << dp[1][n] << endl;    }    return 0;}