'commit'

TangDou/AcWing/FloodFill/1106_dfs_2.cpp

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+#include <bits/stdc++.h>
+using namespace std;
+const int N = 1010;
+
+int n;
+bool f[N][N];
+int h[N][N];
+// 　将两个需要返回的参数，设置为全局变量，则可以正常通过此题。
+//   将两个需要返回的参数，设置为带地址符的变量，则MLE
+bool has_higher, has_lower;
+//	657 ms
+void dfs(int sx, int sy) {
+    f[sx][sy] = true;
+    for (int x = sx - 1; x <= sx + 1; x++) {
+        for (int y = sy - 1; y <= sy + 1; y++) {
+            if (x <= 0 || x > n || y <= 0 || y > n) continue;
+            if (h[sx][sy] != h[x][y]) { // 高度不相等
+                if (h[sx][sy] < h[x][y]) has_higher = true;
+                if (h[sx][sy] > h[x][y]) has_lower = true;
+            } else { // 高度相等
+                if (f[x][y]) continue;
+                dfs(x, y);
+            }
+        }
+    }
+}
+int vally, peak;
+int main() {
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= n; j++)
+            cin >> h[i][j];
+
+    for (int i = 1; i <= n; i++) {
+        for (int j = 1; j <= n; j++) {
+            if (!f[i][j]) {
+                has_higher = has_lower = false;
+                dfs(i, j);
+                if (has_higher && has_lower) continue;
+                if (has_higher) vally++;
+                if (has_lower) peak++;
+            }
+        }
+    }
+
+    // 对于不存在山峰+山谷的一马平地，山峰山谷都输出1
+    if (peak == 0 && vally == 0) peak = 1, vally = 1;
+    printf("%d %d\n", peak, vally);
+    return 0;
+}