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  • BZOJ 1208 set

    思路:

    开俩set

    模拟一下

    就好了

    //By SiriusRen
    #include <bits/stdc++.h>
    using namespace std;
    int n,xx,yy;
    set<int>s1,s2;set<int>::iterator it;
    long long ans;
    int main(){
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d%d",&xx,&yy);
            if(!xx){
                if(s2.empty())s1.insert(yy);
                else{
                    int tmp=2147483647;
                    it=s2.lower_bound(yy);
                    if(it!=s2.begin())it--,tmp=min(tmp,yy-*it),it++;
                    if(it!=s2.end())tmp=min(tmp,*it-yy);
                    if(it!=s2.begin()){
                        it--;
                        if(yy-*it==tmp){s2.erase(it);goto ed1;}
                        it++;
                    }s2.erase(it);
                    ed1:ans+=tmp;
                }
            }
            else{
                if(s1.empty())s2.insert(yy);
                else{
                    int tmp=2147483647;
                    it=s1.lower_bound(yy);
                    if(it!=s1.begin())it--,tmp=min(tmp,yy-*it),it++;
                    if(it!=s1.end())tmp=min(tmp,*it-yy);
                    if(it!=s1.begin()){
                        it--;
                        if(yy-*it==tmp){s1.erase(it);goto ed2;}
                        it++;
                    }s1.erase(it);
                    ed2:ans+=tmp;
                }
            }
        }printf("%lld
    ",ans%1000000);
    }
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  • 原文地址:https://www.cnblogs.com/SiriusRen/p/9371569.html
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