'commit'

TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/6_3.cpp

@@ -0,0 +1,57 @@
+#include <bits/stdc++.h>
+/*
+测试用例：
+5 6
+1 0 1 0 0 0
+0 1 0 1 0 0
+1 0 0 0 1 0
+0 1 0 0 0 1
+1 0 1 0 0 0
+
+答案
+4
+
+
+测试用例II：
+5 6
+1 0 1 0 0 0
+0 1 0 1 1 0
+1 0 0 0 1 0
+0 1 0 0 0 1
+1 0 1 0 0 0
+
+答案
+2
+*/
+using namespace std;
+const int N = 110;
+int a[N][N];
+int s[N][N];
+int res;
+
+int main() {
+    int n, m;
+    cin >> n >> m;
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= m; j++) {
+            cin >> a[i][j];
+            //维护二维前缀和
+            s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + a[i][j];
+        }
+
+    //直接在原数组上递推
+    for (int i = n - 1; i >= 1; i--)        //从倒数第二行开始
+        for (int j = 1; j <= m; j++)        //这个需要考虑一下边界问题
+            if (a[i][j]) {                  //只有当前位置大于0才有考虑的价值
+                a[i][j] += a[i + 1][j + 1]; //递推式,当前位置可以向下扩展的最长对角线长度
+
+                //利用二维前缀和计算区域的总和
+                int x1 = i, y1 = j, x2 = i + a[i][j] - 1, y2 = j + a[i][j] - 1;
+                int cnt = s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1];
+                if (cnt == a[i][j])
+                    res = max(res, a[i][j]); //在递推过程中不断求最大值
+            }
+    //输出
+    cout << res << endl;
+    return 0;
+}