Codeforces 446b dzy loves modification matrix priority queue + structure

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Question:

Given n rows and M columns of matrix K operations, a constant P

Ans = 0;

For each operation

You can select either one row or one column, ANS + = The number and

Then every number in this row (column) is-= P

Largest ans

Ideas:

First, we finally select row I, then the column selects k-I

If we select all rows and then columns,-= I * P is required for each column selection.

In this case, the end is-= I * (k-I) * P

That is, the impact of all rows on columns.

Then we first propose this I * (k-I) * P, so the selection of rows and columns will not affect each other.

You can consider rows and columns separately.

For the case of Row-only:

Pre-process to select 0 times 1 times · maximum value of K rows H [I]

That is, you can run the queue first.

Similarly, if column processing is performed, L [I] indicates that the maximum value of the zero row is obtained.

Then ans = max (H [I] + L [k-I]-I * (k-I) * P)

Note that the result may be a small negative number. ans =-INF and INF must be large enough.

#include <cstdio>#include <algorithm>#include<iostream>#include<string.h>#include <math.h>#include<queue>using namespace std;#define ll long long#define N 1005#define M 1000005ll a[N][N];ll retc[M], retr[M];priority_queue<pair<ll, ll> > sc, sr;int main() {ll n, m, k, p;while(cin>>n>>m>>k>>p){            while(!sc.empty())sc.pop();    while(!sr.empty())sr.pop();        for(ll i = 1; i <= n; i ++) {            for(ll j = 1; j <= m; j ++) {                scanf("%I64d", &a[i][j]);            }        }        for(ll i = 1; i <= n; i ++) {            ll s = 0;            for(ll j = 1; j <= m; j ++) {                s += a[i][j];            }            sr.push(make_pair(s, i));        }        for(ll j = 1; j <= m; j ++) {            ll s = 0;            for(ll i = 1; i <= n; i ++) {                s += a[i][j];            }            sc.push(make_pair(s, j));        }        for(ll i = 1; i <= k; i ++) {            ll s = sr.top().first;            ll id = sr.top().second;            retr[i] = retr[i-1] + s;            s -= p*m;            sr.pop();            sr.push(make_pair(s, id));        }        for(ll i = 1; i <= k; i ++) {            ll s = sc.top().first;            ll id = sc.top().second;            retc[i] = retc[i-1] + s;            s -= p*n;            sc.pop();            sc.push(make_pair(s, id));        }        ll ans = -(ll)(1e18);        for(ll i = 0; i <= k; i ++) {            ans = max(ans, retr[i] + retc[k-i] - (ll)i * (k-i) * p);        }        cout<<ans<<endl;    }return 0;}/*2 2 2 21 32 42 2 5 21 32 43 1 3 13213 2 4 11 11 11 13 1 3 10002 3 3 21 1 -1-1 1 11 5 21 11 2 3 4 5 6*/