Jzoj P4754 矩阵___map+小根堆

版权声明：欢迎借鉴，谢绝抄搬。 https://blog.csdn.net/Gx_Man_VIP/article/details/88647276

题目大意：

分析：

因为权值非负，一个矩阵的权值肯定≥其子矩阵。那么建立一个堆，
开始把所有最小的和谐矩阵加入堆，每次从堆中取出最小的和谐矩阵，同时扩展出最多四个新的和谐矩阵（上下左右多一行或一列），注意判重，重复 $k$次即可。
然后注意一下每个矩阵拓展的时候只需要→↓或者←↑ 2个方向即可，4个方向会炸

代码：

#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <map>
#include <cstdio>
#include <algorithm>

#define JR(i, j) (i-1)*m+j
#define N 1005

using namespace std;

typedef long long ll;

int n, m, mina, minb, K;
ll sum[N][N];

struct Node {
    int x1, y1, x2, y2;
    ll Get_Sum() const {
        return sum[x2][y2] - sum[x1 - 1][y2] -sum[x2][y1 - 1] + sum[x1 - 1][y1 - 1];
    }
    friend bool operator < (Node aa, Node bb) 
    {
         return aa.Get_Sum() > bb.Get_Sum();
    }
    Node(){};
    Node(int a, int b, int c, int d)
    {
         x1 = a, y1 = b, x2 = c, y2 = d;
    }
};
priority_queue <Node> Q;
map <pair<int, int>, int> id;


inline void Insert(Node &A) 
{
    int x1 = A.x1, y1 = A.y1, x2 = A.x2, y2 = A.y2;
    if (x2 < n && !id[make_pair(JR(x1, y1), JR(x2 + 1, y2))]) 
	   Q.push(Node(x1, y1, x2 + 1, y2)), id[make_pair(JR(x1, y1), JR(x2 + 1, y2))] = 1;
    if (y2 < m && !id[make_pair(JR(x1, y1), JR(x2, y2 + 1))]) 
	   Q.push(Node(x1, y1, x2, y2 + 1)), id[make_pair(JR(x1, y1), JR(x2, y2 + 1))] = 1;
}

int main() 
{
    scanf("%d %d %d %d %d", &n, &m, &mina, &minb, &K);
    for (int i = 1; i <= n; i++) 
	    for (int j = 1; j <= m; j++) 
		{
            int x; scanf("%d", &x),
            sum[i][j] = (ll)sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + x;
    }
    for (int i = 1; i <= n - mina + 1; i++) 
	    for (int j = 1; j <= m - minb + 1; j++) Q.push(Node(i, j, i + mina - 1, j + minb - 1));
    for(int i = 1; i < K; i++) 
	{
        if (!Q.size())
        {
        	printf("-1\n");
        	return 0;
		}
        Node u = Q.top(); Q.pop();
        Insert(u);
    }
    if (!Q.size()) printf("-1\n"); else printf("%lld\n", Q.top().Get_Sum());
    return 0;
}