AtCoder ABC 127F Absolute Minima

题目链接：https://atcoder.jp/contests/abc127/tasks/abc127_f

题目大意

　　初始状态下$f(x) = 0$，现在有 2 种模式的询问，第一种以“1 a b”的形式，需要进行操作$f(x) = f(x) + |x - a| + b$；第二种以“2”的形式，求使得 f(x) 取得最小值的 x 取值和 f(x) 值，如果有多个 x，输出任意一个即可。

分析

　　考虑第一种询问已经出现了 k 次，现在遇到第二种询问。此时$f(x) = \sum_{i = 1}^k (|x - a_i| + b_i)$，可以发现$b_i$完全是独立于 x 的，所以只要讨论$\sum_{i = 1}^k |x - a_i|$的最小值即可。

　　于是问题就退化成在数轴上找一个点，使得它到给定的 k 个点的距离之和最短。

　　当点个数为奇数，最优解是中位数。 
　　当点个数为偶数，最优解是位于中间的两个数中任取一个。

　　所以这个问题的本质就是让你动态找中位数，维护中位数可以用双堆。

代码如下

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3
  4 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
  5 #define Rep(i,n) for (int i = 0; i < (n); ++i)
  6 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
  7 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
  8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
  9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
 10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 12
 13 #define pr(x) cout << #x << " = " << x << "  "
 14 #define prln(x) cout << #x << " = " << x << endl
 15
 16 #define LOWBIT(x) ((x)&(-x))
 17
 18 #define ALL(x) x.begin(),x.end()
 19 #define INS(x) inserter(x,x.begin())
 20
 21 #define ms0(a) memset(a,0,sizeof(a))
 22 #define msI(a) memset(a,inf,sizeof(a))
 23 #define msM(a) memset(a,-1,sizeof(a))
 24
 25 #define MP make_pair
 26 #define PB push_back
 27 #define ft first
 28 #define sd second
 29
 30 template<typename T1, typename T2>
 31 istream &operator>>(istream &in, pair<T1, T2> &p) {
 32     in >> p.first >> p.second;
 33     return in;
 34 }
 35
 36 template<typename T>
 37 istream &operator>>(istream &in, vector<T> &v) {
 38     for (auto &x: v)
 39         in >> x;
 40     return in;
 41 }
 42
 43 template<typename T1, typename T2>
 44 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
 45     out << "[" << p.first << ", " << p.second << "]" << "\n";
 46     return out;
 47 }
 48
 49 inline int gc(){
 50     static const int BUF = 1e7;
 51     static char buf[BUF], *bg = buf + BUF, *ed = bg;
 52
 53     if(bg == ed) fread(bg = buf, 1, BUF, stdin);
 54     return *bg++;
 55 }
 56
 57 inline int ri(){
 58     int x = 0, f = 1, c = gc();
 59     for(; c<48||c>57; f = c==‘-‘?-1:f, c=gc());
 60     for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
 61     return x*f;
 62 }
 63
 64 typedef long long LL;
 65 typedef unsigned long long uLL;
 66 typedef pair< double, double > PDD;
 67 typedef pair< int, int > PII;
 68 typedef pair< string, int > PSI;
 69 typedef set< int > SI;
 70 typedef vector< int > VI;
 71 typedef vector< PII > VPII;
 72 typedef map< int, int > MII;
 73 typedef pair< LL, LL > PLL;
 74 typedef vector< LL > VL;
 75 typedef vector< VL > VVL;
 76 typedef priority_queue< int > PQIMax;
 77 typedef priority_queue< int, VI, greater< int > > PQIMin;
 78 const double EPS = 1e-10;
 79 const LL inf = 0x7fffffff;
 80 const LL infLL = 0x7fffffffffffffffLL;
 81 const LL mod = 1e9 + 7;
 82 const int maxN = 2e5 + 7;
 83 const LL ONE = 1;
 84 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 85 const LL oddBits = 0x5555555555555555;
 86
 87
 88 int Q;
 89 PQIMax maxH;
 90 PQIMin minH;
 91 LL sumB, sumLA, sumRA, ans;
 92
 93 int main(){
 94     INIT();
 95     cin >> Q;
 96     while(Q--) {
 97         int x, a, b;
 98         cin >> x;
 99         if(x == 1) {
100             cin >> a >> b;
101             sumB += b;
102             if(maxH.size() == minH.size()) {
103                 if(maxH.empty()) {
104                     maxH.push(a);
105                     sumLA += a;
106                 }
107                 else if(a >= maxH.top()) {
108                     minH.push(a);
109                     sumRA += a;
110                 }
111                 else {
112                     maxH.push(a);
113                     sumLA += a;
114                 }
115             }
116             else if(maxH.size() > minH.size()) {
117                 if(a >= maxH.top()) {
118                     minH.push(a);
119                     sumRA += a;
120                 }
121                 else {
122                     maxH.push(a);
123                     sumLA += a;
124                     sumLA -= maxH.top();
125                     minH.push(maxH.top());
126                     sumRA += maxH.top();
127                     maxH.pop();
128                 }
129             }
130             else {
131                 if(a < minH.top()) {
132                     maxH.push(a);
133                     sumLA += a;
134                 }
135                 else {
136                     minH.push(a);
137                     sumRA += a;
138                     sumRA -= minH.top();
139                     maxH.push(minH.top());
140                     sumLA += minH.top();
141                     minH.pop();
142                 }
143             }
144         }
145         else {
146             int x = maxH.top();
147             ans = sumRA - sumLA + sumB;
148             if(maxH.size() > minH.size()) ans += maxH.top();
149             else if(maxH.size() < minH.size()) {
150                 ans -= minH.top();
151                 x = minH.top();
152             }
153             cout << x << " " << ans << endl;
154         }
155     }
156     return 0;
157 }

原文地址：https://www.cnblogs.com/zaq19970105/p/10941714.html