[Update] dp + 메모리 절약을 위한 로직 적용

Author: rkdehdgns1230

Dynamic_programming/problem solving/[2096]_내려가기.cpp

@@ -0,0 +1,59 @@
+#include <iostream>
+#define MAX 100001
+#define MAX_VAL 900001
+#define MIN_VAL -1
+using namespace std;
+
+int n;
+int dp_max[2][3];
+int dp_min[2][3];
+
+bool inRange(int x) {
+	return (0 <= x && x <= 2);
+}
+
+int main() {
+	ios::sync_with_stdio(false);
+	cin.tie(NULL);
+	cout.tie(NULL);
+
+	cin >> n;
+
+	//memory 초과 문제를 줄여보자.
+	for (int i = 0; i < n; i++) {
+		int x, y, z;
+		cin >> x >> y >> z;
+
+		dp_max[1][0] = (dp_max[0][0] + x > dp_max[0][1] + x) ? dp_max[0][0] + x : dp_max[0][1] + x;
+		
+		dp_max[1][1] = (dp_max[0][0] + y > dp_max[0][1] + y) ? dp_max[0][0] + y : dp_max[0][1] + y;
+		dp_max[1][1] = (dp_max[1][1] > dp_max[0][2] + y) ? dp_max[1][1] : dp_max[0][2] + y;
+
+		dp_max[1][2] = (dp_max[0][1] + z > dp_max[0][2] + z) ? dp_max[0][1] + z : dp_max[0][2] + z;
+
+
+		dp_min[1][0] = (dp_min[0][0] + x < dp_min[0][1] + x) ? dp_min[0][0] + x : dp_min[0][1] + x;
+
+		dp_min[1][1] = (dp_min[0][0] + y < dp_min[0][1] + y) ? dp_min[0][0] + y : dp_min[0][1] + y;
+		dp_min[1][1] = (dp_min[1][1] < dp_min[0][2] + y) ? dp_min[1][1] : dp_min[0][2] + y;
+
+		dp_min[1][2] = (dp_min[0][1] + z < dp_min[0][2] + z) ? dp_min[0][1] + z : dp_min[0][2] + z;
+
+
+		//한 칸 내린다.
+		for (int j = 0; j < 3; j++) {
+			dp_max[0][j] = dp_max[1][j];
+			dp_min[0][j] = dp_min[1][j];
+		}
+	}
+
+	int max_res = MIN_VAL, min_res = MAX_VAL;
+	for (int j = 0; j < 3; j++) {
+		max_res = (max_res > dp_max[1][j]) ? max_res : dp_max[1][j];
+		min_res = (min_res < dp_min[1][j]) ? min_res : dp_min[1][j];
+	}
+
+	cout << max_res << " " << min_res;
+
+	return 0;
+}