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 == ' '); 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 }