feat: 添加P4087题解实现

实现P4087题目解决方案，包括输入处理、排序和统计逻辑。使用map维护奶牛产奶量排名变化，计算需要更新广告牌的次数。

src/10/20/P4087.cpp

@@ -0,0 +1,60 @@
+#include <algorithm>
+#include <cmath>
+#include <cstdint>
+#include <iostream>
+#include <istream>
+#include <iterator>
+#include <map>
+using ll = int64_t;
+
+const ll maxn = 10'0000 + 5, inf = 1e9 + 7;
+struct C {
+    ll d, c, a;
+    inline bool operator<(const C &o) const {
+        return d < o.d;
+    }
+} c[maxn];
+ll n, g, ans;
+std::map<ll, ll> cow_num, num_cowcount;
+
+int main() {
+    std::iostream::sync_with_stdio(false);
+    std::cin.tie(nullptr);
+
+    std::cin >> n >> g;
+    for (ll i = 1; i <= n; i++) {
+        std::cin >> c[i].d >> c[i].c >> c[i].a;
+    }
+    std::sort(c + 1, c + 1 + n);
+
+    num_cowcount[0] = inf;
+
+    for (ll i = 1; i <= n; i++) {
+
+        ll max_num_before = num_cowcount.rbegin()->first;
+        bool just_one_was_max = num_cowcount.rbegin()->second==1;
+
+        ll old_cownum = cow_num[c[i].c];
+
+        bool was_top = (old_cownum == max_num_before);
+
+        num_cowcount[old_cownum]--;
+        if (num_cowcount[old_cownum] == 0) {
+            num_cowcount.erase(old_cownum);
+        }
+
+        ll new_cownum = old_cownum + c[i].a;
+        cow_num[c[i].c] = new_cownum;
+
+        num_cowcount[new_cownum]++;
+
+        ll max_num_after = num_cowcount.rbegin()->first;
+
+        bool is_top = (new_cownum == max_num_after);
+
+        if ((max_num_before != max_num_after && !just_one_was_max) || (was_top != is_top)) {
+            ans++;
+        }
+    }
+    std::cout << ans << "\n";
+}