HDOJ 4302 - Holedox Eating 优先队列

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题意:

Holedox在一个一维的坐标轴上...初始在0点...现在某些个时刻..会在某些位置放置蛋糕..Holedox在一些时刻进食..Holedox进食满足每次找距离最近的..若两头距离一样.按照原路继续走....问最终Holedox走了多长的路...

题解:

优先队列的典型应用....由于每走一步必定不会跨过蛋糕...用两个队列来维护哪些数在当前位置左侧...哪些在右侧....

Program:

#include<iostream>#include<stdio.h>#include<string.h>#include<cmath>#include<queue>#include<stack>#include<set>#include<time.h>#include<map>#include<algorithm>#define ll long long#define eps 1e-5#define oo 1000000007#define pi acos(-1.0)#define MAXN 100005using namespace std;   priority_queue<int> ql;priority_queue<int,vector<int>,greater<int> > qr; //int型直接改变顺序int main(){                       int T,cases,L,n,ans,now,face;;             scanf("%d",&T);       for (cases=1;cases<=T;cases++)       {                 scanf("%d%d",&L,&n);                while (!ql.empty()) ql.pop();                while (!qr.empty()) qr.pop();                now=ans=face=0;                while (n--)                {                        int tp,x;                        scanf("%d",&tp);                        if (!tp)                         {                               scanf("%d",&x);                               if (x>=now) qr.push(x);                                 else ql.push(x);                        }else                        {                               if (!ql.size() && !qr.size()) continue;                               if (!ql.size()) ans+=qr.top()-now,now=qr.top(),face=0,qr.pop();                               else                               if (!qr.size()) ans+=now-ql.top(),now=ql.top(),face=1,ql.pop();                               else                               if (now-ql.top()==qr.top()-now)                               {                                       if (!face) ans+=qr.top()-now,now=qr.top(),qr.pop();                                         else ans+=now-ql.top(),now=ql.top(),ql.pop();                               }                               else                               if (now-ql.top()<qr.top()-now)                                      ans+=now-ql.top(),now=ql.top(),face=1,ql.pop();                                 else ans+=qr.top()-now,now=qr.top(),face=0,qr.pop();                        }                }                printf("Case %d: %d\n",cases,ans);       }         return 0;}