题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3480 Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 999999/400000 K (Java/Others) Little D is really interested in the theorem of sets recently. There’s a problem that confused him a long time.   …

题目链接:hdu 3480 Division 题意: 给你一个有n个数的集合S,现在让你选出m个子集合,使这m个子集合并起来为S,并且每个集合的(max-min)2 之和要最小. 题解: 运用贪心的思想,肯定首先将全部的数排好序,然后设dp[i][j]表示前j个数分为i个集合的最优解. 则有dp[i][j]=min{dp[i-1][k]+(a[j]-a[k+1])2}(0<k<j). 这样写出来是三层for的dp,考虑用斜率优化降维. 假设l<k<j,对于dp[i][j],k到j为…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2829 T. E. Lawrence was a controversial figure during World War I. He was a British officer who served in the Arabian theater and led a group of Arab nationals in guerilla strikes against the Ottoman Emp…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5956 题意:一颗树上每条边有个权值,每个节点都有新闻要送到根节点就是1节点,运送过程中如果不换青蛙就是走过的所有边权之和的平方,如果换就每次更换要加上P,也就是求“每个节点到根节点这段路径切分成几块之后 [每块的权值和的平方加上(块个数-1)*P] 的最小值”.然后找到所有节点中消耗最大的那个是多少. 题解:设 dist[ i ] 表示节点 i 到根节点的距离,有 dp[ i ] = min(dp[…

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:…

Division Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 999999/400000 K (Java/Others) Total Submission(s): 3984    Accepted Submission(s): 1527 Problem Description Little D is really interested in the theorem of sets recently. There’s a pro…

解题思路 第一步显然是将原数组排序嘛--然后分成一些不相交的子集,这样显然最小.重点是怎么分. 首先,我们写出一个最暴力的\(DP\): 我们令$F[ i ][ j ] $ 为到第\(i\)位,分成\(j\)组的代价,我们可以写出如下 $ DP$ for( LL i = 1; i <= N; ++i ) F[ i ][ 1 ] = sqr( A[ i ] - A[ 1 ] ); for( LL j = 2; j <= M; ++j ) for( LL i = j; i <= N; ++i…

题目大意:一个有n个数的集合,现在要求将他分成m+1个子集,对子集i设si表示该集合中最大数与最小数的差的平方.求所有si的和的最小值.n<=10000,m<=5000. 分析:最优解的m个集合肯定不会相交,也不会出现空集,而且每个子集的数必定是连续的. 所以可以将n个数先排序,在来进行dp求解. f[i][j]表示前j个数分成i个集合的最优解. 转移方程为:f[i][j]=min(f[i-1][k]+(num[j]-num[k+1])^2 设决策点k1<k2,若有k2比k1更优,则有:…

Problem Description Little D is really interested in the theorem of sets recently. There’s a problem that confused him a long time.  Let T be a set of integers. Let the MIN be the minimum integer in T and MAX be the maximum, then the cost of set T if…

题目链接:hdu 2829 Lawrence 题意: 在一条直线型的铁路上,每个站点有各自的权重num[i],每一段铁路(边)的权重(题目上说是战略价值什么的好像)是能经过这条边的所有站点的乘积之和..然后给你m个炮弹,让你选择破坏掉m段铁路,使剩下的整条铁路的战略价值最小. 题解: 和hdu 3480 Division(斜率优化DP)这题相同,只是方程不同而已,改改就行了. #include<bits/stdc++.h> #define F(i,a,b) for(int i=a;i<=…