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

 #include <bits/stdc++.h>

 using namespace std;

 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

 #define Rep(i,n) for (int i = 0; i < (n); ++i)

 #define For(i,s,t) for (int i = (s); i <= (t); ++i)

 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)

 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)

 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)

 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)

 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)

 #define pr(x) cout << #x << " = " << x << "  "

 #define prln(x) cout << #x << " = " << x << endl

 #define LOWBIT(x) ((x)&(-x))

 #define ALL(x) x.begin(),x.end()

 #define INS(x) inserter(x,x.begin())

 #define ms0(a) memset(a,0,sizeof(a))

 #define msI(a) memset(a,inf,sizeof(a))

 #define msM(a) memset(a,-1,sizeof(a))

 #define MP make_pair

 #define PB push_back

 #define ft first

 #define sd second

 template<typename T1, typename T2>

 istream &operator>>(istream &in, pair<T1, T2> &p) {

     in >> p.first >> p.second;

     return in;

 }

 template<typename T>

 istream &operator>>(istream &in, vector<T> &v) {

     for (auto &x: v)

         in >> x;

     return in;

 }

 template<typename T1, typename T2>

 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {

     out << "[" << p.first << ", " << p.second << "]" << "\n";

     return out;

 }

 inline int gc(){

     static const int BUF = 1e7;

     static char buf[BUF], *bg = buf + BUF, *ed = bg;

     if(bg == ed) fread(bg = buf, , BUF, stdin);

     return *bg++;

 } 

 inline int ri(){

     int x = , f = , c = gc();

     for(; c<||c>; f = c=='-'?-:f, c=gc());

     for(; c>&&c<; x = x* + c - , c=gc());

     return x*f;

 }

 typedef long long LL;

 typedef unsigned long long uLL;

 typedef pair< double, double > PDD;

 typedef pair< int, int > PII;

 typedef pair< string, int > PSI;

 typedef set< int > SI;

 typedef vector< int > VI;

 typedef vector< PII > VPII;

 typedef map< int, int > MII;

 typedef pair< LL, LL > PLL;

 typedef vector< LL > VL;

 typedef vector< VL > VVL;

 typedef priority_queue< int > PQIMax;

 typedef priority_queue< int, VI, greater< int > > PQIMin;

 const double EPS = 1e-;

 const LL inf = 0x7fffffff;

 const LL infLL = 0x7fffffffffffffffLL;

 const LL mod = 1e9 + ;

 const int maxN = 2e5 + ;

 const LL ONE = ;

 const LL evenBits = 0xaaaaaaaaaaaaaaaa;

 const LL oddBits = 0x5555555555555555;

 int Q;

 PQIMax maxH;

 PQIMin minH;

 LL sumB, sumLA, sumRA, ans;

 int main(){

     INIT();

     cin >> Q;

     while(Q--) {

         int x, a, b;

         cin >> x;

         if(x == ) {

             cin >> a >> b;

             sumB += b;

             if(maxH.size() == minH.size()) {

                 if(maxH.empty()) {

                     maxH.push(a);

                     sumLA += a;

                 }

                 else if(a >= maxH.top()) {

                     minH.push(a);

                     sumRA += a;

                 }

                 else {

                     maxH.push(a);

                     sumLA += a;

                 }

             }

             else if(maxH.size() > minH.size()) {

                 if(a >= maxH.top()) {

                     minH.push(a);

                     sumRA += a;

                 }

                 else {

                     maxH.push(a);

                     sumLA += a;

                     sumLA -= maxH.top();

                     minH.push(maxH.top());

                     sumRA += maxH.top();

                     maxH.pop();

                 }

             }

             else {

                 if(a < minH.top()) {

                     maxH.push(a);

                     sumLA += a;

                 }

                 else {

                     minH.push(a);

                     sumRA += a;

                     sumRA -= minH.top();

                     maxH.push(minH.top());

                     sumLA += minH.top();

                     minH.pop();

                 }

             }

         }

         else {

             int x = maxH.top();

             ans = sumRA - sumLA + sumB;

             if(maxH.size() > minH.size()) ans += maxH.top();

             else if(maxH.size() < minH.size()) {

                 ans -= minH.top();

                 x = minH.top();

             }

             cout << x << " " << ans << endl;

         }

     }

     return ;

 }

巴特西

AtCoder ABC 127F Absolute Minima

题目大意

分析

代码如下

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