【杂题】 codeforces 446B - DZY Loves Modification

官方题解给出的解法是先预处理出行和列取0到k的最大值，然后从0到k枚举，找到最大的r[i]+c[i]-i*(k-i)*p。。。

#include <iostream>  
#include <sstream>  
#include <algorithm>  
#include <vector>  
#include <queue>  
#include <stack>  
#include <map>  
#include <set>  
#include <bitset>  
#include <cstdio>  
#include <cstring>  
#include <cstdlib>  
#include <cmath>  
#include <climits>  
#define maxn 1000005
#define eps 1e-6 
#define mod 10007 
#define INF 99999999  
#define lowbit(x) (x&(-x))  
//#define lson o<<1, L, mid  
//#define rson o<<1 | 1, mid+1, R  
typedef long long LL;
using namespace std;

struct node
{
	LL x;
	bool operator < (const node &a) const {
		return a.x>x;
	}
}tmp;
priority_queue<node> q1, q2;
LL R[1005], C[1005];
LL r[maxn], c[maxn];

int main(void)
{
	int n, m, k, p;
	int i, j;
	LL ans, a;
	scanf("%d%d%d%d", &n, &m, &k, &p);
	for(i = 1; i <= n; i++)
		for(j = 1; j <= m; j++) {
			scanf("%I64d", &a);
			R[i] += a;
			C[j] += a;
		}
	for(i = 1; i <= n; i++) {
		tmp.x = R[i];
		q1.push(tmp);
	}
	for(i = 1; i <= m; i++) {
		tmp.x = C[i];
		q2.push(tmp);
	}
	for(i = 1; i <= k; i++) {
		tmp = q1.top();
		q1.pop();
		r[i] = r[i-1] + tmp.x;
		tmp.x -= m*p;
		q1.push(tmp);
	}
	for(i = 1; i <= k; i++) {
		tmp = q2.top();
		q2.pop();
		c[i] = c[i-1] + tmp.x;
		tmp.x -= n*p;
		q2.push(tmp);
	}
	ans = max(r[k], c[k]);
	for(i = 1; i <= k; i++)
		ans = max(ans, r[i] + c[k-i] - 1ll*(k-i)*i*p);
	printf("%I64d\n", ans);
	return 0;
}