CF446B DZY Loves Modification 优先队列

最新推荐文章于 2019-04-22 14:17:23 发布

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最新推荐文章于 2019-04-22 14:17:23 发布

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原文链接：http://www.cnblogs.com/zxyqzy/p/10215498.html

本文探讨了一个矩阵游戏中的策略问题，目标是在限定操作次数下，通过修改矩阵元素获得最大乐趣值。文章详细介绍了游戏规则，包括对行和列进行操作的具体方式，并提供了一种算法解决方案，通过枚举行和列的操作次数来求解最优解。

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 each element of the row by p. This operation brings to DZY the value of pleasure equal to the sum of elements of the row before the decreasing.
Pick some column of the matrix and decrease each element of the column by p. This operation brings to DZY the value of pleasure equal to the sum of elements of the column before the decreasing.

DZY wants to know: what is the largest total value of pleasure he could get after performing exactly k modifications? Please, help him to calculate this value.

Input

The first line contains four space-separated integers n, m, k and p (1 ≤ n, m ≤ 10³; 1 ≤ k ≤ 10⁶; 1 ≤ p ≤ 100).

Then n lines follow. Each of them contains m integers representing a_ij (1 ≤ a_ij ≤ 10³) — the elements of the current row of the matrix.

Output

Output a single integer — the maximum possible total pleasure value DZY could get.

Examples

Input

Copy

2 2 2 2
1 3
2 4

Output

Copy

Input

Copy

2 2 5 2
1 3
2 4

Output

Copy

Note

For the first sample test, we can modify: column 2, row 2. After that the matrix becomes:


1 1
0 0

For the second sample test, we can modify: column 2, row 2, row 1, column 1, column 2. After that the matrix becomes:


-3 -3
-2 -2
貌似行和列不太好处理？
假设先对行进行处理了 i 次，那么列自然就是 k-i 次处理；
对行操作结束后： sum-=m*p*i;
此时对列的就是 -= (k-i)*p*i;
那么我们枚举行和列的操作次数即可；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 2000005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
	ll x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/


priority_queue<ll>r, c;
ll n, m;
ll maxc[maxn], maxr[maxn];
ll a[2000][2000];

int main() {
	//ios::sync_with_stdio(0);
	ll k, p;
	cin >> n >> m >> k >> p;
	ll sumr = 0, sumc = 0;
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < m; j++) {
			rdllt(a[i][j]);
			sumr += a[i][j];
		}
		r.push(sumr); sumr = 0;
	}
	for (int i = 0; i < m; i++) {
		for (int j = 0; j < n; j++) {
			sumc += a[j][i];
		}
		c.push(sumc); sumc = 0;
	}
	ll maxx = -1e17 - 4;
	for (int i = 1; i <= k; i++) {
		ll tmp = r.top(); r.pop();
		maxr[i] = maxr[i - 1] + tmp;
		r.push(tmp - m * p);
	}
	for (int i = 1; i <= k; i++) {
		ll tmp = c.top(); c.pop();
		maxc[i] = maxc[i - 1] + tmp;
		c.push(tmp - n * p);
	}
	for (int i = 0; i <= k; i++) {
		ll ans = maxc[i] + maxr[k - i] - (ll)i*(k - i)*p;
		maxx = max(maxx, ans);
	}
	cout << maxx << endl;
	return 0;
}

转载于:https://www.cnblogs.com/zxyqzy/p/10215498.html