题目链接：

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 
      
         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
       
         31 istream &operator>>(istream &in, pair
        
          &p) { 32     in >> p.first >> p.second; 33     return in; 34 } 35   36 template
         
           37 istream &operator>>(istream &in, vector
          
            &v) { 38     for (auto &x: v) 39         in >> x; 40     return in; 41 } 42   43 template
           
             44 ostream &operator<<(ostream &out, const std::pair
            
              &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 }