leetcode1847 最近的房间

有n个房间，rooms[i]={id[i],size[i]}表示第i个房间编号为id[i]，面积为size[i]，房间编号唯一。
现有k个查询query[j]={preferred[j],minsize[j]}，表示要求面积至少为minsize[j]，且编号跟preferred[j]最接近，如果有多个，选编号小的那个。
1<=n<=1E5; 1<=k<=1E4; 1<=id[i],preferred[i]<=1E7; 1<=size[i],minsize[i]<=1E7

分析：像这种二维偏序查询问题，一般做法是离线处理，对查询按第一维排序，按顺序加入集合，然后在集合中查询第二维信息，这样把二维降成一维来处理。
对于本题，将查询按面积从大到小排序，每次查询前将满足面积条件的房间加入集合，然后在该集合中寻找编号最接近的元素。

struct node {
    int idx, id, siz, ans;
};
class Solution {
public:
    vector<int> closestRoom(vector<vector<int>>& rooms, vector<vector<int>>& queries) {
        int n = rooms.size();
        int q = queries.size();
        sort(rooms.begin(), rooms.end(), [&](auto &x, auto &y){return x[1]>y[1];});
        vector<node> vn(q);
        for (int i = 0; i < q; i++) {
            vn[i] = {i, queries[i][0], queries[i][1], 0};
        }
        sort(vn.begin(), vn.end(), [&](auto &x, auto &y){return x.siz > y.siz;});
 
        vector<int> ans(q);
        set<int> st;
        for (int i = 0, j = 0; i < q; i++) {
            while (j < n && rooms[j][1] >= vn[i].siz) {
                st.insert(rooms[j][0]);
                j += 1;
            }
            int best = -1, diff = INT_MAX;
            auto it = st.lower_bound(vn[i].id);
            if (it != st.end() && abs(*it - vn[i].id) < diff) {
                diff = abs(*it - vn[i].id);
                best = *it;
            }
            if (it != st.begin()) {
                --it;
                if (abs(*it - vn[i].id) <= diff) {
                    diff = abs(*it - vn[i].id);
                    best = *it;
                }
            }
            ans[vn[i].idx] = best;
        }
        return ans;
    }
};