From ea4d0774312a8409b1fcb618b2d54f8651714d51 Mon Sep 17 00:00:00 2001
From: TooYoungTooSimp <6648049+TooYoungTooSimp@users.noreply.github.com>
Date: Tue, 12 Feb 2019 17:38:08 +0800
Subject: [PATCH] Tue, 12 Feb 2019 17:38:08 GMT

---
 C279991/.tutorial.md | 10 ++++++++++
 C279991/W.cpp        | 14 ++++++++++++++
 2 files changed, 24 insertions(+)

diff --git a/C279991/.tutorial.md b/C279991/.tutorial.md
index 2024cb0..0f844e8 100644
--- a/C279991/.tutorial.md
+++ b/C279991/.tutorial.md
@@ -107,3 +107,13 @@ multiset的count好像格外的慢呢。或许是通过`distance(lower_bound,upp
 看起来像个什么离散化后最短路一样的东西。实际上map套map就可以了。
 
 //（如果强行说正解是树套树会不会显得高级一点？）
+
+### W - Glass Carving
+
+跟上面那个U差不多，只是xy轴要分开维护。
+
+正版题解上还有一种线性时间复杂度的离线算法，把正序切割改成倒序合并即可。
+
+### End
+
+那么这一堆内容就结束了。
diff --git a/C279991/W.cpp b/C279991/W.cpp
index 7c92ea4..ae763c7 100644
--- a/C279991/W.cpp
+++ b/C279991/W.cpp
@@ -4,5 +4,19 @@
 using namespace std;
 int main()
 {
+    int w, h, n, p;
+    scanf("%d%d%d", &w, &h, &n);
+    set<int64_t> _s[2] = {{0, w}, {0, h}};
+    multiset<int64_t> _ms[2] = {{w}, {h}};
+    for (char op[3]; n--; printf("%lld\n", *_ms[0].rbegin() * *_ms[1].rbegin()))
+    {
+        scanf("%s%d", op, &p);
+        auto &s = _s[*op == 'H'];
+        auto &ms = _ms[*op == 'H'];
+        auto it = s.upper_bound(p);
+        auto len1 = *it - p, len2 = p - *--it;
+        ms.erase(ms.find(len1 + len2));
+        s.insert(p), ms.insert(len1), ms.insert(len2);
+    }
     return 0;
 }