zoj 1196 Fast Food 动态规划

//dp[i][k]表示前i个店添加k个供应点所达到的最小值
//状态转移方程为：dp[i][k] = min(dp[j][k-1], dis[j+1][i]),
//其中k-1 <= j <= i-1， dis[i][j]表示从第i个饭店到第j个饭店添加一个供应点所达到的最小值，取i,j中间值即可
//初始化:dp[0][0] = 0;
#include <iostream>
#include <cstring>
using namespace std;

const int N = 205;
const int INF = 1000000000;

int dp[N][N];
int dis[N][N];
int num[N];

int n, m;

int my_abs(int a)
{
	if (a > 0)
		return a;
	return -a;
}

int main()
{
	int cases = 1;
	while (cin >> n >> m && (n || m))
	{
		for (int i = 1; i <= n; i++)
			cin >> num[i];

		memset(dis, 0, sizeof(dis));
		for (int i = 1; i <= n; i++)
		{
			for (int j = i; j <= n; j++)
			{
				int mid = (i+j)/2;
				for (int k = i; k <= j; k++)
					dis[i][j] += my_abs(num[mid] - num[k]);
			}
		}

		for (int i = 0; i <= n; i++)
			for (int j = 0; j <= n; j++)
				dp[i][j] = INF;
		
		dp[0][0] = 0;

		for (int i = 1; i <= n; i++)
			for (int k = 1; k <= m; k++)		
				for (int j = k-1; j < i; j++)
					dp[i][k] = min(dp[j][k-1] + dis[j+1][i], dp[i][k]);
			
		cout << "Chain " << cases++ << endl;
		cout << "Total distance sum = " << dp[n][m] << endl;
		cout << endl;
	}
	return 0;
}