uva 10304 Optimal Binary Search Tree

题目大意：给出N个结点（已知每个结点的权值）来建树，建树时要满足以下规则：左子树的节点的值要全小于父节点，右子树的节点的值要全大于父节点。要求最后建出的树总权值最小。总权值=各结点乘以层数（从0层开始）之后相加的和。

解题思路：$dp[i][j]$代表区间第i个结点到第j个结点组成的树最小的总权值。$dp[j][i] = min(dp[j][i], dp[j][k - 1] + dp[k + 1][i] + sum[i] - sum[j - 1] - num[k])$

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
using namespace std;
typedef long long ll;
int main() {

    return 0;
}
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#define N 300
using namespace std;
typedef long long ll;
int num[N];
int dp[N][N], sum[N];
int main() {
    int n;
    while (scanf("%d", &n) == 1) {
        sum[0] = 0;
        for (int i = 1; i <= n; i++) {
            scanf("%d", &num[i]);
            sum[i] = sum[i - 1] + num[i];   
        }
        memset(dp, 0, sizeof(dp));
        for (int i = 2; i <= n; i++) {
            for (int j = i - 1; j > 0; j--) {
                dp[j][i] = 0xFFFFFFF;
                for (int k = j; k <= i; k++) {
                    dp[j][i] = min(dp[j][i], dp[j][k - 1] + dp[k + 1][i] + sum[i] - sum[j - 1] - num[k]);  
                }
            }
        }
        printf("%d\n", dp[1][n]);
    }
    return 0;
}