AcWing 106 动态中位数(对顶堆)

题目链接

解题思路

  对顶堆模板题，一个堆对应一半。

代码

priority_queue<int, vector<int>, less<int> > l, clears1;
priority_queue<int, vector<int>, greater<int> > r, clears2;
vector<int> ans;
int main(){
    int t; scanf("%d", &t);
    while(t--) {
        int num, m; scanf("%d%d",&num,&m);
        for (int i = 1; i<=m; ++i) {
            int tmp; scanf("%d",&tmp);
            //大根堆存[l,mid] 小根堆存[mid+1,r]
            if(r.empty()||tmp>r.top()) r.push(tmp);
            else l.push(tmp);
            //维护两个堆的大小
            while (l.size()>i/2) {
                r.push(l.top());
                l.pop();
            }
            while(l.size()<i/2) {
                l.push(r.top());
                r.pop();
            }
            if (i&1) ans.push_back(r.top());
        }
        printf("%d %d\n", num, (m+1)/2);
        int sz = ans.size();
        for (int i = 0; i<sz; ++i) printf("%d%c", ans[i], (i+1)%10==0?'\n':' ');
        if (sz%10) putchar(endl);
        ans.clear(); l = clears1; r = clears2;
    }
    return 0;
}