超时算法,利用2的特殊性,用2个multiset来维护。单个multiset维护没法立即找到中位数。

其实也可以只用1个multiset,用一个中位指针,++,--来维护中位数。

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<queue>
#include<vector>
#include<string>
#include<map>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
const int maxn = 1e4 + 7;
int n;
multiset<int> up, lo;
stack<int> sta;
void adapt() {
    int cnt = up.size() + lo.size();
    int upsz = ceil(cnt / 2.0);
    while (up.size() < upsz) {
        up.insert(*lo.begin());
        lo.erase(lo.begin());
    }
    while (up.size() > upsz) {
        int x = *up.rbegin();
        lo.insert(x);
        up.erase(up.find(x));
    }
}
void push(int x) {
    up.insert(x);
    adapt();
}
int peek() {
    return *(up.rbegin());
}
void pop(int x) {
    if (up.find(x) != up.end())
        up.erase(up.find(x));
    else
        lo.erase(lo.find(x));
    adapt();
}
int main() {
    freopen("in.txt", "r", stdin);
    cin >> n;
    while (sta.empty() == false)
        sta.pop();
    up.clear(), lo.clear();
    char cmd[13];
    int x;
    while (n--) {
        cin >> cmd;
        if (cmd[1] != 'u') {
            if (lo.empty() && up.empty()) {
                puts("Invalid");
                continue;
            }
            if (cmd[1] == 'o') {
                int x = sta.top();
                sta.pop();
                cout << x << endl;
                pop(x);
            } else if (cmd[1] == 'e') {
                cout << peek() << endl;
            }
        } else {
            cin >> x;
            sta.push(x);
            push(x);
        }
    }
    return 0;
}

此题正解树状数组

#include<stdio.h>
#include<cstring>
#include<iostream>
#include<string>
using namespace std;

;
int c[N];

int lowbit(int i){
    return i&(-i);
}

void add(int pos,int value){
    while(pos<N){
        c[pos]+=value;
        pos+=lowbit(pos);
    }
}

int sum(int pos){
    ;
    ){
        res+=c[pos];
        pos-=lowbit(pos);
    }
    return res;
}

int find(int value){
    ,r=N-,median,res;
    ){
        ==)
            median=(l+r)/;
        else
            median=(l+r-)/;
        res=sum(median);
        if(res<value)
            l=median;
        else
            r=median;

    }
    ;
}

int main(){
    //freopen("D://test.txt","r",stdin);
    ];
    ,n,pos;
    memset(c,,sizeof(c));
    scanf("%d",&n);
    while(n--){
        scanf("%s",ss);
        ]=='u'){
            scanf("%d",&pos);
            stack[++top]=pos;
            add(pos,);
        }]=='o'){
            ){
                printf("Invalid\n");
                continue;
            }
            int out=stack[top];
            add();
            printf("%d\n",stack[top--]);
        }]=='e'){
            ){
                printf("Invalid\n");
                continue;
            }
            int res;
            ==)
                res=find(top/);
            else
                res=find((top+)/);
            printf("%d\n",res);

        }else{
            printf("Invalid\n");
        }
    }

    ;
}  

类似题目:zoj3612

05-11 13:25