HDU 4302 Holedox Eating （STL + 模拟）

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

题意：在一条直线上一开始一只小动物在0点，然后有两种操作，一种是在某一点上放一块蛋糕，另一种是去吃一块蛋糕。其中吃蛋糕有规定，先吃离小动物最近的蛋糕，如果有两块蛋糕离小动物的距离一样近，先吃当时的朝向的蛋糕，问所有操作以后小动物所走的路多长。

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

using namespace std;

int L,n;
priority_queue<int,vector<int>,greater<int> >q1;    //升序，（比loc大的，即蛋糕在自己位置的右边）
priority_queue<int> q2;     //降序，（比loc小的，即蛋糕在自己位置的左边）
int loc,dir,ans=;  //loc代表自己的位置，dir==0表示向左，dir==1表示向右，ans为答案

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

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

int main(){

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

    int t,cases=;
    scanf("%d",&t);
    while(t--){
        while(!q1.empty())
            q1.pop();
        while(!q2.empty())
            q2.pop();
        scanf("%d%d",&L,&n);
        loc=,dir=,ans=;
        int op,x;
        while(n--){
            scanf("%d",&op);
            if(op==){
                scanf("%d",&x);
                if(x>=loc)
                    q1.push(x);
                else
                    q2.push(x);
            }else{
                if(q1.empty() && q2.empty())    //没有蛋糕则什么也不做
                    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==)
                            Forward();
                        else
                            Back();
                    }
                }
            }
        }
        printf("Case %d: %d\n",++cases,ans);
    }
    return ;
}