HDU 4302 Holedox Eating(STL+シミュレーション)

Holedox Eating
Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2862    Accepted Submission(s): 952
Problem Description
Holedox is a small animal which can be considered as one point. It lives in a straight pipe whose length is L. Holedox can only move along the pipe. Cakes may appear anywhere in the pipe, from time to time. When Holedox wants to eat cakes, it always goes to the nearest one and eats it. If there are many pieces of cake in different directions Holedox can choose, Holedox will choose one in the direction which is the direction of its last movement. If there are no cakes present, Holedox just stays where it is.
 
 
Input
The input consists of several test cases. The first line of the input contains a single integer T (1 <= T <= 10), the number of test cases, followed by the input data for each test case.The first line of each case contains two integers L,n(1<=L,n<=100000), representing the length of the pipe, and the number of events. 
The next n lines, each line describes an event. 0 x(0<=x<=L, x is a integer) represents a piece of cake appears in the x position; 1 represent Holedox wants to eat a cake.
In each case, Holedox always starts off at the position 0.
 
 
Output
Output the total distance Holedox will move. Holedox don’t need to return to the position 0.
 
 
Sample Input
3 10 8 0 1 0 5 1 0 2 0 0 1 1 1 10 7 0 1 0 5 1 0 2 0 0 1 1 10 8 0 1 0 1 0 5 1 0 2 0 0 1 1
 
 
Sample Output
Case 1: 9 Case 2: 4 Case 3: 2
 
 
Author
BUPT
 
 
Source
2012 Multi-University Training Contest 1
 
 
Recommend
zhuyuanchen520
 
 
标题:一直線上で最初は1匹の小動物が0時で、それから2つの操作があります.1つはある点にケーキを置いて、もう1つはケーキを食べに行きます.その中でケーキを食べる規定があって、まず小動物に一番近いケーキを食べて、もし2つのケーキが小動物の距離に同じように近いならば、まずその時の向きのケーキを食べて、すべての操作の後で小動物が歩いている道がどれだけ長いかを聞きます.
　

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>

using namespace std;

int L,n;
priority_queue<int,vector<int>,greater<int> >q1;    //  ，（ loc  ，           ）
priority_queue<int> q2;     //  ，（ loc  ，           ）
int loc,dir,ans=0;  //loc       ，dir==0    ，dir==1    ，ans   

void Forward(){     //   
    int tmp=q1.top();
    q1.pop();
    ans+=tmp-loc;
    loc=tmp;
    dir=1;
}

void Back(){    //   
    int tmp=q2.top();
    q2.pop();
    ans+=loc-tmp;
    loc=tmp;
    dir=0;
}

int main(){

    //freopen("input.txt","r",stdin);

    int t,cases=0;
    scanf("%d",&t);
    while(t--){
        while(!q1.empty())
            q1.pop();
        while(!q2.empty())
            q2.pop();
        scanf("%d%d",&L,&n);
        loc=0,dir=1,ans=0;
        int op,x;
        while(n--){
            scanf("%d",&op);
            if(op==0){
                scanf("%d",&x);
                if(x>=loc)
                    q1.push(x);
                else
                    q2.push(x);
            }else{
                if(q1.empty() && q2.empty())    //          
                    continue;
                if(q1.empty() && !q2.empty()){
                    Back();
                    continue;
                }
                if(!q1.empty() && q2.empty()){
                    Forward();
                    continue;
                }
                if(!q1.empty() && !q2.empty()){
                    int tmp1=q1.top();
                    int tmp2=q2.top();
                    if(tmp1-loc>loc-tmp2)
                        Back();
                    else if(tmp1-loc<loc-tmp2)
                        Forward();
                    else{
                        if(dir==1)
                            Forward();
                        else
                            Back();
                    }
                }
            }
        }
        printf("Case %d: %d
",++cases,ans);
    }
    return 0;
}