很明顯的區間DP，d(i, j)表示取第i堆到第j堆石子最少花費，轉移方程d(i, j) = min(d(i, k) + d(k+1, j)) +  sum[j] - sum[i].

       四邊形a <= b < c <= d，有w(b, c) <= w(a, d)滿足決策單調性 .  且w(a, c) + w(b, d) <= w(b, c) + w(a, d)滿足四邊形不等式 .

AC程式碼：

#include<cstdio>
#include<algorithm>
using namespace std;
const int inf = 1 << 30;
const int maxn = 1000 + 5;
int dp[maxn][maxn], s[maxn][maxn], a[maxn], sum[maxn];

//四邊形不等式優化 
int solve(int n){
	for(int i = 1; i <= n; ++i) {
		dp[i][i] = 0;
		s[i][i] = i;
	}
	for(int l = 2; l <= n; ++l) {
		for(int i = 1; i <= n - l + 1; ++i) {
			int j = i + l - 1;
			dp[i][j] = inf;
			int x = s[i][j-1], y = s[i+1][j];
			for(int k = x; k <= y; ++k) {
				int h = dp[i][k] + dp[k+1][j] + sum[j] - sum[i-1];
				if(h < dp[i][j]) {
					dp[i][j] = h;
					s[i][j] = k;
				}
			}
		}
	}
	return dp[1][n];
}

int main(){
	int n;
	while(scanf("%d", &n) == 1){
		sum[0] = 0;
		for(int i = 1; i <= n; ++i) {
			scanf("%d", &a[i]);
			sum[i] = sum[i - 1] + a[i];
		}
		printf("%d\n", solve(n));
	}
	return 0;
}