24-04-29 비숍

yuyu0830/README.md

@@ -8,4 +8,5 @@
 | 4차시 | 2024.03.26 |  분할정복  | [행렬 제곱](https://www.acmicpc.net/problem/10830)  | - |
 | 5차시 | 2024.03.29 |  탐색  | [스도쿠](https://www.acmicpc.net/problem/2239)  | - |
 | 6차시 | 2024.04.03 |  자료구조  | [나만 안되는 연애](https://www.acmicpc.net/problem/14621)  | - |
+| 10차시| 2024.04.29 |  백트래킹  | [비숍](https://www.acmicpc.net/problem/1799)  | - |
 ---

yuyu0830/탐색/1799.cpp

@@ -0,0 +1,68 @@
+#include <iostream>
+#include <vector>
+
+using namespace std;
+
+typedef pair<int, int> ii;
+
+int leftArr[2][11] = {0, };
+int rightArr[2][11] = {0, };
+int map[11][11] = {0, };
+
+int n, ans[2] = {0, }, m[2] = {0, };
+vector<ii> v[2];
+
+void solve(int color, int idx) {
+    // current color, idx bishop's position
+    int x = v[color][idx].first;
+    int y = v[color][idx].second;
+
+    // position about each diagonal direction
+    int rightPos = (x + y) / 2;
+    int leftPos = ((n - x - 1) + y) / 2;
+
+    // if can set bishop in (x, y) position
+    if (map[x][y] && !rightArr[color][rightPos] && !leftArr[color][leftPos]) {
+        // then check diagonal position
+        rightArr[color][rightPos] = 1;
+        leftArr[color][leftPos] = 1;
+
+        // update max value
+        m[color] = max(m[color], ++ans[color]);
+
+        // search same color, another position bishop
+        for (int i = idx + 1; i < v[color].size(); i++) 
+            solve(color, i);
+
+        // set off current bishop
+        rightArr[color][rightPos] = 0;
+        leftArr[color][leftPos] = 0;
+
+        ans[color]--;
+    }
+}
+
+int main() {
+    cin >> n;
+
+    for (int i = 0; i < n; i++) {
+        for (int j = 0; j < n; j++) {
+            int t; cin >> t;
+            map[i][j] = t;
+            // if bishop can set in (i, j) push in vector[board color]
+            if (t) v[i % 2 != j % 2].push_back(make_pair(i, j));
+        }
+    }
+
+    for (int color = 0; color < 2; color++) {
+        for (int i = 0; i < v[color].size(); i++) {
+            // search every position that can set in bishop
+            solve(color, i);
+
+            // if (left bishop + current max value) less then max value then break this search
+            if (ans[color] + (v[color].size() - i - 1) < m[color]) break;
+        }   
+    }
+
+    printf("%d\n", m[0] + m[1]);
+}