hdu-1421 搬寝室

题目链接：hdu-1421

dp[i][j]表示从前j个中取i对，排序后，肯定两两相邻的平方差和最小，若取i，则肯定i与i-1一对，若不选，则取前面值，两者中取最小值，dp[i][j] = min(dp[i][j - 1], dp[i - 1][j - 2] + (s[j] - s[j - 1])*(s[j] - s[j - 1]));

#include<iostream>#include<cstring>#include<algorithm>using namespace std;int dp[1005][2005];// i表示k对，j表示前n个，前j个中取k对消耗的最少疲劳 int maxx=0x3fffffff;//最大值 int main(){    int n,k;while (cin >> n >> k){int i, j;int s[2005];    memset(dp,0,sizeof(dp));//将全部变为0，包括第0行     memset(s,0,sizeof(s));for (int i = 1; i <= n; i++)cin >> s[i];sort(s+1, s + n+1);//从小到大排序，abcd中，肯定ab的平方差加上cd的平方差和最小 for (int i = 1; i <=k; i++)for (int j = 0; j <= n; j++)dp[i][j] = maxx;//除第一行以外全部变为最大值 for(int i=1;i<=k;i++)for (int j = 2*i; j <=n; j++)//i对，则j从2*i开始打表 {//若选这个，则dp[i][j]= dp[i - 1][j - 2] + (s[j] - s[j - 1])*(s[j] - s[j - 1]),不选则取左边的值，两者取最大值 dp[i][j] = min(dp[i][j - 1], dp[i - 1][j - 2] + (s[j] - s[j - 1])*(s[j] - s[j - 1]));}cout << dp[k][n] << endl;}return 0;}