As we know, DZY loves playing games. One day DZY decided to play with a n × m matrix. To be more precise, he decided to modify the matrix with exactly k operations. Each modification is one of the following: Pick some row of the matrix and decrease e…

B. DZY Loves Modification 题目连接: http://www.codeforces.com/contest/446/problem/B Description As we know, DZY loves playing games. One day DZY decided to play with a n × m matrix. To be more precise, he decided to modify the matrix with exactly k opera…

D. DZY Loves Modification time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output As we know, DZY loves playing games. One day DZY decided to play with a n × m matrix. To be more precise, he decid…

题目传送门 题目大意:给一个 \(n*m\) 的矩阵,并进行 \(k\) 次操作,每次操作将矩阵的一行或一列的所有元素的值减 \(p\) ,得到的分数为这次修改之前这一列/一行的元素和,求分数最大值. 我开始的意识流想法是用一个优先队列维护,先把所有元素插入,然后\(k\)次每次取出堆顶,减去乘\(p\)的什么东西,再塞回队中.但是...行与列是会互相影响的鸭,那么怎么搞呢?然后就不会了hh. 正解是努力打破了行与列之间的相互影响,把行与列分开计算.我们考虑行与列是怎么互相影响的:选择\(i\)…

D. DZY Loves Modification time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output As we know, DZY loves playing games. One day DZY decided to play with a n × m matrix. To be more precise, he decid…

D - DZY Loves Modification As we know, DZY loves playing games. One day DZY decided to play with a n × mmatrix. To be more precise, he decided to modify the matrix with exactly koperations. Each modification is one of the following: Pick some row of…

447D - DZY Loves Modification 思路:将行和列分开考虑.用优先队列,计算出行操作i次的幸福值r[i],再计算出列操作i次的幸福值c[i].然后将行取i次操作和列取k-i次操作,那么多加的幸福值就是i*(k-i)*p,因为无论先操作行还是列,每操作一次一个格子只减一次p.这样记录下最大的幸福值. 代码: #include<bits/stdc++.h> using namespace std; #define ll long long ; ; const ll INF=…

B. DZY Loves Modification time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output As we know, DZY loves playing games. One day DZY decided to play with a n × m matrix. To be more precise, he decid…

枚举行取了多少次,如行取了i次,列就取了k-i次,假设行列单独贪心考虑然后相加,那么有i*(k-i)个交点是多出来的:dpr[i]+dpc[k-i]-i*(k-i)*p 枚举i取最大值.... B. DZY Loves Modification time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output As we know, DZY l…

题意:给你一个矩阵,每次选某一行或者某一列,得到的价值为那一行或列的和,然后该行每个元素减去p.问连续取k次能得到的最大总价值为多少. 解法: 如果p=0,即永远不减数,那么最优肯定是取每行或每列那个最大的取k次,所以最优解由此推出. 如果不管p,先拿,最后再减去那些行列交叉点,因为每个点的值只能取一次,而交叉点的值被加了两次,所以要减掉1次,如果取行A次,取列B次,那么最后答案为: res = dp1[A] + dp2[B] - B*(k-A)*p,可以细细体会一下后面那部分. 其中: dp1…