HDOJ-1421 搬寝室

用dp(i)(j)表示前i个物品中取j对的最小疲劳度,状态转移方程: dp[i][j] = min(dp[i-1][j], dp[i-2][j-1] + (num[i]-num[i-1]) * (num[i]-num[i-1]));

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <climits>

using namespace std;

int dp[2002][1002];
int num[2002];
int main()
{
// freopen("in.txt", "r", stdin);
int n, k;

while(cin >> n >> k)
{
for(int i = 1; i <= n; i++)
cin >> num[i];
sort(num+1, num+n+1);
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n/2; j++)
dp[i][j] = INT_MAX;
for(int i = 0; i <= n; i++)
dp[i][0] = 0;

for(int i = 2; i <= n; i++)
for(int j = 1; j <= i/2; j++)
dp[i][j] = min(dp[i-1][j], dp[i-2][j-1] + (num[i]-num[i-1]) * (num[i]-num[i-1]));
cout << dp
[k] << endl;
}
return 0;
}