047.扫描线

扫描线

对边界排序，按顺序扫描，过程中动态维护当前位置状态，从而高效处理区间问题

经典扫描线问题，天际线

leetcode 218

class Solution {
public:vector<vector<int>> getSkyline(vector<vector<int>>& buildings) {set<int>st;for(auto x:buildings){st.insert(x[0]);st.insert(x[1]);}priority_queue<pair<int,int>>pq;// first :: 高度// second :: 区间右端点（区间结束时间）int i=0;vector<vector<int>>ans;int pre=-1,h;for(auto cur:st){// cur :: 当前处理的线while(i<(int)buildings.size()&&buildings[i][0]<=cur){//包含当前扫描线的区间入堆pq.push({buildings[i][2],buildings[i][1]-1});i++;}while(pq.size()&&pq.top().second<cur)pq.pop();//堆顶过期则弹出h=pq.size()?pq.top().first:0;if(h!=pre){ans.push_back({cur,h});}pre=h;}return ans;}
};

luogu P1904

#include<bits/stdc++.h>using namespace std;
typedef pair<int,int> pii;void solve(){int L,H,R;vector<vector<int>>a;while(cin>>L>>H>>R){a.push_back({L,R,H});}sort(a.begin(),a.end(),[](auto &x,auto &y){return x[0]<y[0];});set<int>st;for(auto x:a){st.insert(x[0]);st.insert(x[1]);}priority_queue<pii>pq;int i=0;int pre=-1,h;for(auto cur:st){while(i<(int)a.size()&&a[i][0]<=cur){pq.push({a[i][2],a[i][1]-1});++i;}while(pq.size()&&pq.top().second<cur)pq.pop();h=pq.size()?pq.top().first:0;if(h!=pre){cout<<cur<<' '<<h<<' ';}pre=h;}
}
int main(void){cin.tie(0)->sync_with_stdio(0);int T=1;//cin>>T;while(T--)solve();
}

类似问题

leetcode 1851

class Solution {
public:vector<int> minInterval(vector<vector<int>>& intervals, vector<int>& queries) {vector<pair<int,int>>Q;int n=queries.size();for(int i=0;i<n;++i){Q.push_back({queries[i],i});}ranges::sort(Q);sort(intervals.begin(),intervals.end(),[](auto a,auto b){return a[0]<b[0];});priority_queue<pair<int,int>,vector<pair<int,int>>,greater<>>pq;int i=0;vector<int>ans(n);for(auto [cur,id]:Q){while(i<(int)intervals.size()&&intervals[i][0]<=cur){pq.push({intervals[i][1]-intervals[i][0]+1,intervals[i][1]});++i;}while(pq.size()&&pq.top().second<cur)pq.pop();ans[id]=pq.size()?pq.top().first:-1;}return ans;}
};