区间DP HDU4283

HDU 4283
这题柯南 柯南 柯南
我也不是很懂
大佬题解

//MADE BY Y_is_sunshine;
//#include <bits/stdc++.h>
//#include <memory.h>
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <cstdio>
#include <vector>
#include <string>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>

#define INF 0x3f3f3f3f
#define MAXN 105

const int mod = 1e9 + 7;
const double PI = acos(-1);

using namespace std;

int d[MAXN];
int sum[MAXN];
int dp[MAXN][MAXN];

int main()
{
	freopen("data.txt", "r", stdin);

	int T;
	cin >> T;
	int cnt = 0;
	while (T--) {
		int N;
		scanf("%d", &N);
		for (int i = 1; i <= N; i++)
			scanf("%d", &d[i]), sum[i] = sum[i - 1] + d[i];
		memset(dp, 0, sizeof(dp));
		for (int i = 1; i <= N; i++)
			for (int j = i + 1; j <= N; j++)
				dp[i][j] = INF;
		for (int i = 1; i <= N; i++)
			dp[i][i] = 0;

		for (int len = 1; len <= N; len++) {
			for (int i = 1; i + len - 1 <= N; i++) {
				int j = i + len - 1;
				//dp[i][i] = 0;
				for (int k = i; k <= j; k++)			//这里要k <= j枚举从第1个到最后一个
					//dp[i][j] = min(dp[i][j], dp[i + 1][k] + dp[k + 1][j] + (k - i) * d[i] + (k - i + 1) * (sum[j] - sum[k]));
					dp[i][j] = min(dp[i][j], dp[i + 1][k] + dp[k + 1][j] + (k - i) * d[i] + (k - i + 1) * (sum[j] - sum[k]));
					//上面的状态转移方程是dp[i + 1][k] 而不是 dp[i][k]
			}
		}
		printf("Case #%d: %d\n", ++cnt, dp[1][N]);
		
	}
	

	

	freopen("CON", "r", stdin);
	system("pause");
	return 0;
}