[Epper] 이진경(온라인)

Epper/1971039_등굣길.cpp

@@ -0,0 +1,30 @@
+#include <string>
+#include <vector>
+
+using namespace std;
+
+int solution(int m, int n, vector<vector<int>> puddles) {
+    vector<vector<int>> dp(n + 1, vector<int>(m + 1, 0));
+    int answer = 0;
+
+    for (int i = 0; i < puddles.size(); i++) {
+        int x = puddles[i][1]; //열
+        int y = puddles[i][0]; //행
+        dp[x][y] = -1;
+    }
+
+    dp[1][1] = 1;
+    for (int i = 1; i <= n; i++) {
+        for (int j = 1; j <= m; j++) {
+            if (i == 1 && j == 1)
+                continue;
+            if (dp[i][j] == -1)
+                dp[i][j] = 0;
+            else
+                dp[i][j] = (dp[i - 1][j] + dp[i][j - 1]) % 1000000007;
+        }
+    }
+
+    answer = dp[n][m] % 1000000007;
+    return answer;
+}

Epper/1971039_정수 삼각형.cpp

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+#include <string>
+#include <vector>
+
+using namespace std;
+
+int dp[501][501]; //전역변수니까 0으로 초기화 됨
+
+int solution(vector<vector<int>> triangle) {
+    int answer = 0;
+
+    dp[0][0] = triangle[0][0];
+    for (int i = 0; i < triangle.size(); i++) {
+        for (int j = 0; j < triangle[i].size(); j++) {
+            if (j == 0)
+                dp[i][j] = dp[i - 1][j] + triangle[i][j];
+            else if (j == i)
+                dp[i][j] = dp[i - 1][j - 1] + triangle[i][j];
+            else
+                dp[i][j] = max(dp[i - 1][j - 1], dp[i - 1][j]) + triangle[i][j];
+            dp[i][j] = max(dp[i - 1][j - 1], dp[i - 1][j]) + triangle[i][j];
+            answer = max(answer, dp[i][j]);
+        }
+    }
+
+    return answer;
+}