题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3045 It’s summer vocation now. After tedious milking, cows are tired and wish to take a holiday. So Farmer Carolina considers having a picnic beside the river. But there is a problem, not all the cows co…

题目链接:hdu 3045 Picnic Cows 题意: 有n个奶牛分别有对应的兴趣值,现在对奶牛分组,每组成员不少于t, 在每组中所有的成员兴趣值要减少到一致,问总共最少需要减少的兴趣值是多少. 题解: 分析:先对n个数进行排序,则可以分析出分组成员一定是连续的 dp[i]表示前i个数得到的最少值则:从j~i作为一组 dp[i]=dp[j-1]+sum[i]-sum[j-1]-(i-j+1)*s[j];//sum[i]表示前i个数的和=>dp[i]=dp[j-1]+sum[i]-sum[j-…

Picnic Cows Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2192    Accepted Submission(s): 675 Problem Description It’s summer vocation now. After tedious milking, cows are tired and wish to t…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3045 题目大意:有n个数,可以把n个数分成若干组,每组不得小于m个数,每组的价值=除了该组最小值以外每个值-最小值之和,求使得所有组的价值之和的最小值. 解题思路:将n个数按从小到大排序,处理前i为前缀和为sum[i],则可得出状态转移方程:dp[i]=min{dp[j]+sum[i]-sum[j+1]-a[j+1]*(i-j-1)}(0<=j<i-m+1),再用斜率DP优化即可. 注意:一定要判…

$dp$,斜率优化. 设$dp[i]$表示$1$至$i$位置的最小费用,则$dp[i]=min(dp[j]+s[i]-s[j]-(i-j)*x[j+1])$,$dp[n]$为答案. 然后斜率优化就可以了. 得到了两个教训: ①如果可以从$dp[0]$推过来,那么队列中一开始就压入$0$,不要忘记了. ②$check2$中,要压入哪个位置就判断哪个位置. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<…

题目链接:https://vjudge.net/problem/HDU-3045 Picnic Cows Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3491    Accepted Submission(s): 1124 Problem Description It’s summer vocation now. After tedi…

斜率DP 设dp[i][j]表示前i点,炸掉j条边的最小值.j<i dp[i][j]=min{dp[k][j-1]+cost[k+1][i]} 又由得出cost[1][i]=cost[1][k]+cost[k+1][i]+sum[k]*(sum[i]-sum[k]) cost[k+1][i]=cost[1][i]-cost[1][k]-sum[k]*(sum[i]-sum[k]) 代入DP方程 可以得出 y=dp[k][j-1]-cost[1][k]+sum[k]^2 x=sum[k]. 斜率s…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3507 Zero has an old printer that doesn't work well sometimes. As it is antique, he still like to use it to print articles. But it is too old to work for a long time and it will certainly wear and tear,…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3480 题目大意:将n个数字分成m段,每段价值为(该段最大值-该段最小值)^2,求最小的总价值. 解题思路:很单纯的斜率优化DP,得出状态转移方程:dp[i][j]=min{dp[k][j-1]+(a[i]-a[k+1])^2}(j-1<=k<i),然后斜率优化降到O(n^2)就好了. 注意:数据类型建议用int,不要用long long,后者乘法计算时间是前者的四倍,否则C++可能会超时. 代码:…

        ID Origin Title   20 / 60 Problem A HDU 3507 Print Article   13 / 19 Problem B HDU 2829 Lawrence   1 / 5 Problem C HDU 4528 小明系列故事――捉迷藏   5 / 6 Problem D HDU 1300 Pearls   0 / 42 Problem E HDU 2993 MAX Average Problem   1 / 20 Problem F UVALi…