UVa 10360 Rat Attack (枚举&优化)

来源：互联网发布：第一岛链知乎编辑：程序博客网时间：2024/06/03 15:07

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;}