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 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 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…

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…

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…

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…

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…

题意: k次操作  每次选择一行或一列  得到所选数字的和  并将所选数字同一时候减去p  问最多得到多少 思路: 重点在消除行列间的相互影响 因为每选一行全部列所相应的和都会-p  那么假设选了i次行  则列会-i*p  同理选列 那么影响就能够这样表示 -p*i*(k-i)  把影响提出来  这样行列就不影响了 对于行或列  单独处理时相当于一维的东西  贪心就可以 代码: #include<cstdio> #include<cstring> #include<algor…