76 lines
2.2 KiB
Markdown
76 lines
2.2 KiB
Markdown
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## 1、激光炸弹
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#### 录入训练
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https://www.acwing.com/problem/content/101/
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#### 温故知新
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- 点坐标如何修改为数组方格坐标
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- 二维前缀和
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- 不能开两个数组，因为每个数组是$5000x5000$这么大，就是$5000x5000x4bytes，5000x5000x4/1024/1024=95MB$，这题的空间是$168MB$，所以两个数组会$MLE$。
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一个数组就行了，在自己身上求前缀和。
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- $xy$坐标是从$0$开始的，二维前缀和需要坐标从$(1,1)$开始，需要$x++,y++$
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- 炸弹的范围会超过$r>M$，会导致计算二维前缀和的两层循环失效，造成结果错误。正确的作法是时刻小心观察，循环什么情况下不会执行，然后加上特判。
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- 只枚举右下角即可完成所有边长为$r$的正方形枚举
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### 蓝桥杯第13届省赛中级组第6题
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```cpp {.line-numbers}
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#include <bits/stdc++.h>
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/*
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测试用例I：
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5 6
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1 0 1 0 0 0
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0 1 0 1 0 0
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1 0 0 0 1 0
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0 1 0 0 0 1
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1 0 1 0 0 0
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答案
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4
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测试用例II：
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5 6
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1 0 1 0 0 0
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0 1 0 1 1 0
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1 0 0 0 1 0
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0 1 0 0 0 1
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1 0 1 0 0 0
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答案
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2
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*/
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using namespace std;
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const int INF = 0x3f3f3f3f;
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const int N = 110;
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int a[N][N];
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int s[N][N];
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int res;
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int main() {
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int n, m;
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cin >> n >> m;
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= m; j++) {
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cin >> a[i][j];
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s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + a[i][j];
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}
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for (int i = 1; i <= n; i++) // 100
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for (int j = 1; j <= m; j++) { // 100
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for (int k = 0;; k++) {
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if (i + k > n || j + k > m) break; // 越界
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if (a[i + k][j + k] == 0) break;
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int x1 = i, y1 = j, x2 = i + k, y2 = j + k;
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int cnt = s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1];
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if (cnt != k + 1) break;
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res = max(res, cnt);
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}
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}
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printf("%d\n", res);
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return 0;
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}
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```
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