BNUOJ-26475 Cookie Selection 堆,线段树等

  题目链接:http://www.bnuoj.com/bnuoj/problem_show.php?pid=26475

  题意:每次输入一个操作,如果是数字,那么放入一个容器中,如果是#号,取出当前容器中的中间值。。

  数据结构基础题,显然维护两个堆就可以了,两个堆的size大小不超过1...

  其实各种数据结构都可以搞,比如线段树,先离线然后离散,然后线段树记录size,二分查找。。

 1 //STATUS:C++_AC_856MS_3192KB
 2 #include <functional>
 3 #include <algorithm>
 4 #include <iostream>
 5 //#include <ext/rope>
 6 #include <fstream>
 7 #include <sstream>
 8 #include <iomanip>
 9 #include <numeric>
10 #include <cstring>
11 #include <cassert>
12 #include <cstdio>
13 #include <string>
14 #include <vector>
15 #include <bitset>
16 #include <queue>
17 #include <stack>
18 #include <cmath>
19 #include <ctime>
20 #include <list>
21 #include <set>
22 //#include <map>
23 using namespace std;
24 //#pragma comment(linker,"/STACK:102400000,102400000")
25 //using namespace __gnu_cxx;
26 //define
27 #define pii pair<int,int>
28 #define mem(a,b) memset(a,b,sizeof(a))
29 #define lson l,mid,rt<<1
30 #define rson mid+1,r,rt<<1|1
31 #define PI acos(-1.0)
32 //typedef
33 typedef long long LL;
34 typedef unsigned long long ULL;
35 //const
36 const int N=600010;
37 const int INF=0x3f3f3f3f;
38 const int MOD=1e9+7,STA=8000010;
39 //const LL LNF=1LL<<60;
40 const double EPS=1e-8;
41 const double OO=1e15;
42 const int dx[4]={-1,0,1,0};
43 const int dy[4]={0,1,0,-1};
44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
45 //Daily Use ...
46 inline int sign(double x){return (x>EPS)-(x<-EPS);}
47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
50 template<class T> inline T Min(T a,T b){return a<b?a:b;}
51 template<class T> inline T Max(T a,T b){return a>b?a:b;}
52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
56 //End
57 
58 struct cmp{
59     bool operator()(const int a,const int b){
60         return a<b;
61     }
62 };
63 priority_queue<int,vector<int>,cmp > q1;
64 priority_queue<int,vector<int>,greater<int> > q2;
65 
66 int main()
67 {
68  //   freopen("in.txt","r",stdin);
69     int i,j,a,cnt=0;
70     char s[20];
71     q1.push(-INF);q2.push(INF);
72     while(~scanf("%s",s))
73     {
74         if(s[0]=='#'){
75             if(cnt){
76                 printf("%d
",q1.top());
77                 q1.pop();
78             }
79             else {
80                 printf("%d
",q2.top());
81                 q2.pop();
82             }
83             cnt^=1;
84         }
85         else {
86             sscanf(s,"%d",&a);
87             q1.push(a);
88             cnt^=1;
89         }
90         while(q1.size()>q2.size()){
91             q2.push(q1.top());q1.pop();
92         }
93         while(q1.size()<q2.size()){
94             q1.push(q2.top());q2.pop();
95         }
96     }
97     return 0;
98 }
原文地址:https://www.cnblogs.com/zhsl/p/3282012.html