부록 E 27501번 풀이 추가

Appendix E/solutions/27501.cpp

@@ -1,11 +1,73 @@
-// Authored by : BaaaaaaaaaaarkingDog
+// Authored by : uhwan0723
 // Co-authored by : -
-// http://boj.kr/****************
+// http://boj.kr/7386ea6f37f84fa7a16dcd892d847b97
 #include <bits/stdc++.h>
 using namespace std;
 
-int main(void){
+enum COLOR{
+  R, G, B
+};
+
+const int mn = 5e5+5;
+int n, d[mn][3], beauty[mn][3], res[mn];
+vector<int> adj[mn];
+
+int dfs(int par, int cur, int color){
+  if(d[cur][color] != -1) return d[cur][color];
+  d[cur][color] = 0;
+  for(int nxt : adj[cur]){
+    if(nxt == par) continue;
+    int mx = 0;
+    for(int c = R; c <= B; ++c){
+      if(c == color) continue;
+      mx = max(mx, dfs(cur, nxt, c));
+    }
+    d[cur][color] += mx;
+  }
+  return d[cur][color] += beauty[cur][color];
+}
+
+void btrace(int par, int cur, int color){
+  res[cur] = color;
+  for(int nxt : adj[cur]){
+    if(nxt == par) continue;
+    int mx = 0, col = 0;
+    for(int c = R; c <= B; ++c){
+      if(c == color || d[nxt][c] <= mx) continue;
+      mx = d[nxt][c];
+      col = c;
+    }
+    btrace(cur, nxt, col);
+  }
+}
+
+int main(){
   ios::sync_with_stdio(0);
   cin.tie(0);
+
+  cin >> n;
+  for(int i = 0; i < n-1; ++i){
+    int u, v;
+    cin >> u >> v;
+    adj[u].push_back(v);
+    adj[v].push_back(u);
+  }
+  for(int i = 1; i <= n; ++i)
+    cin >> beauty[i][R] >> beauty[i][G] >> beauty[i][B];
+
+  for(int i = 1; i <= n; ++i)
+    fill(d[i], d[i]+3, -1);
+
+  int mx = 0, start_col = 0;
+  for(int c = R; c <= B; ++c){
+    if(dfs(-1, 1, c) <= mx) continue;
+    mx = d[1][c];
+    start_col = c;
+  }
+  btrace(-1, 1, start_col);
 
-}
+  char col[3] = {'R', 'G', 'B'};
+  cout << mx << '\n';
+  for(int i = 1; i <= n; ++i)
+    cout << col[res[i]];
+}