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 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: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<…

转自PomeCat: "DP的斜率优化--对不必要的状态量进行抛弃,对不优的状态量进行搁置,使得在常数时间内找到最优解成为可能.斜率优化依靠的是数形结合的思想,通过将每个阶段和状态的答案反映在坐标系上寻找解答的单调性,来在一个单调的答案(下标)队列中O(1)得到最优解." https://wenku.baidu.com/view/b97cd22d0066f5335a8121a3.html "一些试题中繁杂的代数关系身后往往隐藏着丰富的几何背景,而借助背景图形的性质,可以使那些…

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…

题意:将n个数分成若干组,每组数字的个数不少于t个,要把每组的数字减小到这组最小值,求所有数字减少的最小值. 先将这n个数从小到大排个序,可以想到一组里面的数一定是排序后相邻的. 设d(i)表示前i个数分完组以后减少的最小值,考虑j~i为一组,则有状态转移方程 还是一样的处理方法,设k < j ≤ i - t,且j~i为一组的值比k~i为一组的值更优. 则有不等式: 化简,把i分离出来,整理成斜率的形式: 写到这里就应该很清楚地能够看出来X和Y的表达式了. #include <iostream…

Description N (2 <= N <= 8,000) cows have unique brands in the range 1..N. In a spectacular display of poor judgment, they visited the neighborhood 'watering hole' and drank a few too many beers before dinner. When it was time to line up for their e…

题面 HDU vjudge 题解 将权值排序,则分组一定是连续的 设$f[i]$表示前$i$头牛的最小代价,则($a[i]$为$i$的权值): $$ f[i] = f[j - 1] + sum[i] - sum[j - 1] - (i - j + 1) * a[j] $$ 套上斜率优化的板子即可. #include<cstdio> #include<cstring> #include<cctype> #include<algorithm> #define R…