diff --git a/TangDou/Topic/PrefixAndSuffix/P1719_1.cpp b/TangDou/Topic/PrefixAndSuffix/P1719_1.cpp
index 5f79f3b..e7fc354 100644
--- a/TangDou/Topic/PrefixAndSuffix/P1719_1.cpp
+++ b/TangDou/Topic/PrefixAndSuffix/P1719_1.cpp
@@ -5,6 +5,7 @@ int n;
 int a[N][N]; // 存储题目中的矩阵
 int s[N][N]; // 二维前缀和
 
+// O(N^4)算法
 int main() {
     cin >> n;
     for (int i = 1; i <= n; i++) {
diff --git a/TangDou/Topic/PrefixAndSuffix/P1719_2.cpp b/TangDou/Topic/PrefixAndSuffix/P1719_2.cpp
index 761b2c1..c21836a 100644
--- a/TangDou/Topic/PrefixAndSuffix/P1719_2.cpp
+++ b/TangDou/Topic/PrefixAndSuffix/P1719_2.cpp
@@ -1,26 +1,31 @@
 #include <bits/stdc++.h>
 using namespace std;
-const int N = 130;
-int a[N][N], s[N][N];
+
+const int N = 330;
+int n;
+int a[N][N], dp[N];
+int res = INT_MIN;
+
 int main() {
-    int n, res = INT_MIN;
     cin >> n;
-    for (int i = 1; i <= n; i++)
+    // 前缀和(竖直方向)
+    for (int i = 1; i <= n; i++) {
         for (int j = 1; j <= n; j++) {
             cin >> a[i][j];
-            s[i][j] = s[i - 1][j] + a[i][j]; // 利用一维前缀和，把当前第j列前面的1~j-1列的值，累加和压缩到第j列，每列都是如此处理
+            a[i][j] += a[i - 1][j];
         }
-
-    // O(N^3)算法
-    for (int i = 1; i <= n; i++) {
-        for (int j = 0; j < i; j++) { // 前缀和需要下标从0开始
-            int sum = 0;
+    }
+    // 降维变成一维dp
+    for (int i = 0; i <= n - 1; i++) {
+        for (int j = i + 1; j <= n; j++) {
             for (int k = 1; k <= n; k++) {
-                sum = max(sum, 0) + s[i][k] - s[j][k];
-                res = max(res, sum);
+                dp[k] = max(a[j][k] - a[i][k], dp[k - 1] + a[j][k] - a[i][k]);
+                res = max(res, dp[k]);
             }
         }
     }
-    cout << res << endl;
+
+    cout << res;
+
     return 0;
 }
diff --git a/TangDou/Topic/【前缀和与差分】题单.md b/TangDou/Topic/【前缀和与差分】题单.md
index 5f0c5b6..5951421 100644
--- a/TangDou/Topic/【前缀和与差分】题单.md
+++ b/TangDou/Topic/【前缀和与差分】题单.md
@@ -274,7 +274,41 @@ signed main() {
 
 #### [$P1719$ 最大加权矩阵](https://www.luogu.com.cn/problem/P1719)
 
-**解题思路**
+**$O(N^4)$算法**
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+using namespace std;
+const int N = 130;
+int n;
+int a[N][N]; // 存储题目中的矩阵
+int s[N][N]; // 二维前缀和
+
+int main() {
+    cin >> n;
+    for (int i = 1; i <= n; i++) {
+        for (int j = 1; j <= n; j++) {
+            cin >> a[i][j];
+            s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + a[i][j];
+        }
+    }
+
+    int mx = INT_MIN; // 存储答案
+    // 遍历左上角坐标，与，右下角坐标
+    for (int x1 = 1; x1 <= n; x1++) {
+        for (int y1 = 1; y1 <= n; y1++) {
+            for (int x2 = 1; x2 <= n; x2++)
+                for (int y2 = 1; y2 <= n; y2++) {
+                    if (x2 < x1 || y2 < y1) continue;                                            // 如果左上角比右下角还要大，就不用求了，下一个
+                    mx = max(mx, s[x2][y2] + s[x1 - 1][y1 - 1] - s[x2][y1 - 1] - s[x1 - 1][y2]); // 求最大值
+                }
+        }
+    }
+    cout << mx << endl; // 输出
+    return 0;
+}
+```
+
+**$O(N^3)$算法**
 这道题和$P1115$最大子段和思路类似。
 只是将一维升到二维,很重要的解决思路:**矩阵压缩**。
 如果能把二维降到一维，那就好处理了。