hihocoder 1634 ACM-ICPC北京赛区2017网络同步赛H题 Puzzle Game （DP最大子矩阵和）

题目链接：http://hihocoder.com/problemset/problem/1634?sid=1477014

题意：给一个n*m的矩阵，其最大子矩阵为矩阵内所有值的和最大的一个子矩阵，现给一个值p，从矩阵中任选一个值替换为p或者不操作，问所有可能操作结果矩阵的最大子矩阵和的最小值为多少

思路：（u[i],d[i], l[i], r[i]）四个数组分别代表第i行（列）向上（向左、向下、向右）矩阵的最大子矩阵和，可以用dp在O()的复杂度下预处理四个数组的值，具体dp方法：https://blog.youkuaiyun.com/AmarisCR/article/details/88869976

然后暴力n*m扫整个矩阵，若a[i][j]<=p则替换无效，否则更新答案为

                                                  max(MAX - a[i][j]+p,u[i-1],d[i+1],l[j-1],r[j+1])

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

typedef long long ll;
typedef double lb;
const ll mod = 1e9 + 7;
const int maxn = 150 + 50;
const ll inf = 0x3f3f3f3f;

ll u[maxn], d[maxn], l[maxn], r[maxn];
int n, m, p;
ll a[maxn][maxn];
ll res[maxn], col[maxn], MAX;

ll cmp(ll x, ll y, ll z, ll k)
{
	ll tmp = max(x, y);
	tmp = max(tmp, z);
	return max(tmp, k);
}

int main()
{
	while (~scanf("%d%d%d", &n, &m, &p))
	{
		MAX = -inf;
		memset(u, -inf, sizeof(u));
		memset(d, -inf, sizeof(d));
		memset(l, -inf, sizeof(l));
		memset(r, -inf, sizeof(r));
		for (int i = 1; i <= n; i++)
		{
			for (int j = 1; j <= m; j++)
			{
				scanf("%lld", &a[i][j]);
			}
		}
		for (int i = n; i >= 1; i--)
		{
			memset(res, 0, sizeof(res));
			d[i] = max(d[i], d[i + 1]);
			for (int j = i; j <= n; j++)
			{
				for (int k = 1; k <= m; k++)
				{
					res[k] += a[j][k];
				}
				ll tr = -inf, tmp = 0;
				for (int k = 1; k <= m; k++)
				{
					if (tmp > 0)
					{
						tmp += res[k];
					}
					else
					{
						tmp = res[k];
					}
					tr = max(tmp, tr);
				}
				d[i] = max(d[i], tr);
				MAX = max(tr, MAX);
			}
		}
		for (int i = m; i >= 1; i--)
		{
			memset(res, 0, sizeof(res));
			r[i] = max(r[i], r[i + 1]);
			for (int j = i; j <= m; j++)
			{
				for (int k = 1; k <= n; k++)
				{
					res[k] += a[k][j];
				}
				ll tr = -inf, tmp = 0;
				for (int k = 1; k <= n; k++)
				{
					if (tmp > 0)
					{
						tmp += res[k];
					}
					else
					{
						tmp = res[k];
					}
					tr = max(tmp, tr);
				}
				r[i] = max(r[i], tr);
				MAX = max(tr, MAX);
			}
		}
		for (int i = 1; i <= n; i++)
		{
			memset(res, 0, sizeof(res));
			u[i] = max(u[i], u[i - 1]);
			for (int j = i; j > 0; j-- )
			{
				for (int k = 1; k <= m; k++)
				{
					res[k] += a[j][k];
				}
				ll tr = -inf, tmp = 0;
				for (int k = 1; k <= m; k++)
				{
					if (tmp > 0)
					{
						tmp += res[k];
					}
					else
					{
						tmp = res[k];
					}
					tr = max(tmp, tr);
				}
				u[i] = max(u[i], tr);
				MAX = max(tr, MAX);
			}
		}
		for (int i = 1; i <= m; i++)
		{
			memset(res, 0, sizeof(res));
			l[i] = max(l[i], l[i - 1]);
			for (int j = i; j > 0; j--)
			{
				for (int k = 1; k <= n; k++)
				{
					res[k] += a[k][j];
				}
				ll tr = -inf, tmp = 0;
				for (int k = 1; k <= n; k++)
				{
					if (tmp > 0)
					{
						tmp += res[k];
					}
					else
					{
						tmp = res[k];
					}
					tr = max(tmp, tr);
				}
				l[i] = max(l[i], tr);
				MAX = max(tr, MAX);
			}
		}
		ll ans = MAX;
		for (int i = 1; i <= n; i++)
		{
			for (int j = 1; j <= m; j++)
			{
				if (a[i][j] <= p)continue;
				ll x, y, z, k;
				x = u[i - 1];
				y = d[i + 1];
				z = l[j - 1];
				k = r[j + 1];
				ll tmp = cmp(x, y, z, k);
				tmp = max(tmp, MAX - a[i][j] + p);
				ans = min(ans,tmp);
			}
		}
		printf("%lld\n", ans);
	}
}