HDU 4302 Holedox Eating (两个优先队列)

比赛时队友尝试用一个优先队列做，后来发现用两个更合适。

第一次用priority_queue。。。

每次在吃cake时尽量不改变方向，吃最近的cake，每次pop之后并不改变左右队列。

菜鸟忘了刷新队列WA了好几次。。。。教训啊T——T

HDU 4302

Holedox Eating

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

//#pragma comment(linker, "/STACK:102400000,102400000")
#include "iostream"
#include "cstring"
#include "algorithm"
#include "cmath"
#include "cstdio"
#include "sstream"
#include "queue"
#include "vector"
#include "string"
#include "stack"
#include "cstdlib"
#include "deque"
#include "fstream"
#include "map"
using namespace std;
typedef long long LL;
const int INF = 0x1fffffff;
const int MAXN = 1000000+100;
#define eps 1e-14
const int mod = 95041567;

int main()
{
    //freopen("in","r",stdin);
    //freopen("out","w",stdout);
    int L,n,pos,t,dir,dis;
    scanf("%d",&t);
    for (int kase=1; kase<=t; kase++)
    {
        dis=0;
        pos=0;
        dir=1;
        priority_queue<int ,vector<int> ,greater<int> > right;//之前放在main里了。。。深刻教训。。。。
        priority_queue<int> left;
        scanf("%d%d",&L,&n);//L似乎没什么用

        for (int i=0; i<n; i++)
        {
            int a,b;
            scanf("%d",&a);
            if (a==0)
            {
                scanf("%d",&b);
                if (b<=pos) left.push(b);//把小的和在一个位置的都放在左边，后面处理时加一个判断
                else right.push(b);
            }
            else if (a==1)
            {
                if (!left.empty()&&!right.empty())
                {
                    int posl=left.top();
                    int posr=right.top();
                    if (pos-posl==posr-pos)
                    {
                        if (dir==1)//判断方向
                        {
                            dis+=posr-pos;
                            pos=posr;
                            right.pop();
                        }
                        else
                        {
                            dis+=pos-posl;
                            pos=posl;
                            left.pop();
                        }

                    }
                    else if (pos-posl>posr-pos)
                    {
                        dis+=posr-pos;
                        pos=posr;
                        dir=1;
                        right.pop();
                    }
                    else if (pos-posl<posr-pos)
                    {
                        if (posl!=pos)//特判
                        {
                            dis+=pos-posl;
                            pos=posl;
                            dir=-1;
                        }
                        left.pop();
                    }
                }
                else if (!left.empty())
                {
                    if (left.top()!=pos)//同上
                    {
                        dis+=pos-left.top();
                        pos=left.top();
                        dir=-1;//改变方向
                    }
                    left.pop();
                }
                else if (!right.empty())
                {
                    dis+=right.top()-pos;
                    pos=right.top();
                    dir=1;
                    right.pop();
                }

            }
        }
        printf("Case %d: %d\n",kase,dis);
    }
}