http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=1301


由于格子数远大于鼠窝,所以以鼠窝为出发点,增加鼠窝周围d范围内的“格子的值”,最后扫描每个格子输出最大格子值即可。


完整代码:

/*0.078s*/

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1030;

int cnt[MAXN][MAXN];

int main()
{
	int T, d, n, i, j, p, q;
	int x, y, num, lmax;
	int x_min, x_max, y_min, y_max;
	scanf("%d", &T);
	while (T--)
	{
		scanf("%d%d", &d, &n);
		memset(cnt, 0, sizeof(cnt));
		for (i = 0; i < n; ++i)
		{
			scanf("%d%d%d", &x, &y, &num);
			x_min = max(0, x - d), x_max = min(1024, x + d);
			y_min = max(0, y - d), y_max = min(1024, y + d);
			for (p = x_min; p <= x_max; ++p)
				for (q = y_min; q <= y_max; ++q)
					cnt[p][q] += num;
		}
		lmax = x = y = 0;
		for (i = 0; i < 1025; ++i)
			for (j = 0; j < 1025; ++j)
				if (cnt[i][j] > lmax)
					x = i, y = j, lmax = cnt[i][j];
		printf("%d %d %d\n", x, y, lmax);
	}
	return 0;
}