【dp】you are the one

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4283

题解： 当最优解下， a1在j的位置排出， 则a2 ——aj-1 和 aj——an为两个独立事件，

　　状态转移方程：

 dp[i][i + j] = min(dp[i][i + j], dp[i + 1][i + k] + k *arr[i] + dp[i + k + 1][i + j] +( k + 1) *sum[i + k + 1][i + j]) 做法和取石子类似。

 1 /***Good Luck***/
 2 #define _CRT_SECURE_NO_WARNINGS
 3 #include <iostream>
 4 #include <cstdio>
 5 #include <cstdlib>
 6 #include <cstring>
 7 #include <string>
 8 #include <algorithm>
 9 #include <stack>
10 #include <map>
11 #include <queue>
12 #include <vector>
13 #include <set>
14 #include <functional>
15 #include <cmath>
16 #include <numeric>
17
18 #define Zero(a) memset(a, 0, sizeof(a))
19 #define Neg(a)  memset(a, -1, sizeof(a))
20 #define All(a) a.begin(), a.end()
21 #define PB push_back
22 #define inf 0x3f3f3f3f
23 #define inf2 0x7fffffffffffffff
24 #define ll long long
25 using namespace std;
26 //#pragma comment(linker, "/STACK:102400000,102400000")
27 void get_val(int &a) {
28     int value = 0, s = 1;
29     char c;
30     while ((c = getchar()) == ‘ ‘ || c == ‘\n‘);
31     if (c == ‘-‘) s = -s; else value = c - 48;
32     while ((c = getchar()) >= ‘0‘ && c <= ‘9‘)
33         value = value * 10 + c - 48;
34     a = s * value;
35 }
36 const int maxn = 104;
37 int arr[maxn];
38 int n;
39 int dp[maxn][maxn];
40 int sum[maxn][maxn];
41
42 int main() {
43     //freopen("data.out", "w", stdout);
44     //freopen("data.in", "r", stdin);
45     //cin.sync_with_stdio(false);
46     int T;
47     cin >> T;
48     for (int ii = 1; ii <= T; ++ii) {
49         cin >> n;
50         for (int i = 1; i <= n; ++i) scanf("%d", arr + i);
51         Zero(dp);
52         Zero(sum);
53         for (int i = 1; i <= n; ++i) {
54             int t = 0;
55             for (int j = 0; j + i <= n; ++j) {
56                 t += arr[i + j];
57                 sum[i][i + j] = t;
58             }
59         }
60         for (int j = 1; j <= n - 1; ++j) {
61             for (int i = 1; i + j <= n; ++i) {
62                 dp[i][i + j] = (1 << 31) - 1;
63                 for (int k = 0; k <= j; ++k) {
64                     dp[i][i + j] = min(dp[i][i + j], dp[i + 1][i + k] + k *arr[i] + dp[i + k + 1][i + j] +( k + 1) *sum[i + k + 1][i + j]);
65                 }
66             }
67         }
68         printf("Case #%d: ", ii);
69         cout << dp[1][n] << endl;
70     }
71     return 0;
72 }