?
搬寝室
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20642 Accepted Submission(s): 7013
?
Problem Description
2 1 1 3Sample Output
4题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1421
题目大意：久违的中文题，不解释
题目分析：显然左右手物品重量越接近越好，先按照重量从小到大排序，然后预处理出相邻两个数的差值的平方b[i] = (a[i + 1] - a[i])^2，这样预处理也直接把2k化成k了，显然b[i]和b[i + 1]是不能同时取的，定义dp[i][j]为b数组的前i个取了j个时的最少疲劳度，转移方程dp[i][j] = min(dp[i - 1][j], dp[i - 2][j - 1] + b[i])，初始化dp[i][0]=0，dp[1][1]=b[1]，其他INF
#include#include #include using namespace std; int const MAX = 2005; int const INF = 0x3fffffff; int dp[MAX][MAX], a[MAX], b[MAX]; int main() { int n, k; while(scanf(%d %d, &n, &k) != EOF) { for(int i = 1; i <= n; i++) scanf(%d, &a[i]); sort(a + 1, a + n + 1); int sum = 0; for(int i = 1; i < n; i++) b[i] = (a[i + 1] - a[i]) * (a[i + 1] - a[i]); n --; for(int i = 0; i <= n; i++) { for(int j = 0; j <= k; j++) dp[i][j] = INF; dp[i][0] = 0; } dp[1][1] = b[1]; for(int i = 2; i <= n; i++) for(int j = 1; j <= k; j++) dp[i][j] = min(dp[i - 1][j], dp[i - 2][j - 1] + b[i]); printf(%d , dp[n][k]); } }
?
?