P2004. 领地选择

考点

  • 前缀和

题解

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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e3 + 50;
int n, m, c, ans_x, ans_y, pre[maxn][maxn];
ll ans = LLONG_MIN;

ll query(int x1, int y1, int x2, int y2) {
return pre[x2][y2] - pre[x1 - 1][y2] - pre[x2][y1 - 1] + pre[x1 - 1][y1 - 1];
}

int main() {
cin >> n >> m >> c;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
cin >> pre[i][j];
pre[i][j] += pre[i - 1][j] + pre[i][j - 1] - pre[i - 1][j - 1];
}
}
for (int i = 1; i <= n - c + 1; ++i) {
ll sum = 0;
int l = 1, r = 1;
while (l <= r && r <= n) {
if (r - l + 1 > c) sum -= query(i, l, i + c - 1, l), ++l;
sum += query(i, r, i + c - 1, r);
if (r >= c && sum > ans) ans = sum, ans_x = i, ans_y = l;
++r;
}
}
cout << ans_x << " " << ans_y;
return 0;
}

思路

枚举每个c * c的正方形和即可,这里的枚举方式选择对每行执行一次滑动窗口;

正方形和选择用二维前缀和进行优化。