貪吃蛇

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

思路：開始嘗試使用數組，超時，然後考慮各種數據結構，沒錯，就是使用優先隊列，一下子就AC了。。

//聽說網上還有2種做法，一定要去了解一下

#include <iostream>
#include 
 
<cstring>
#include <algorithm>
#include <queue>
using namespace std;

int main()
{
    int t;
    cin >> t;
    for(int T = 1; T <= t; T++)
    {
        priority_queue<int> s;
        priority_queue<int, vector<int>, greater<int> > ss;
        int she = 0;
        long long sum = 0;
        int l, n;
        int fangxiang = 1;//1為右，0為左
        cin >> l >> n;
        while(n--)
        {
            int c;
            cin >> c;
            if(c == 0)
            {
                int x;
                cin >> x;
                if(x > she)ss.push(x);
                else s.push(x);
            }
            else
            {


                if(!s.empty() && !ss.empty() && ss.top() - she > she - s.top()){sum += she - s.top();she = s.top();fangxiang = 0;s.pop();continue;}
                if(s.empty() && !ss.empty()){sum += ss.top() - she;she = ss.top();fangxiang = 1;ss.pop();continue;}
                if(!s.empty() && !ss.empty() &&ss.top() - she < she - s.top()){sum += ss.top() - she;she = ss.top();fangxiang = 1;ss.pop();continue;}
                if(!s.empty() && ss.empty()){sum += she - s.top();she = s.top();fangxiang = 0;s.pop();continue;}
                if(!s.empty() && !ss.empty() && ss.top() - she == she - s.top() )
                {
                    if(fangxiang == 1)
                    {
                        sum += ss.top() - she;
                        she = ss.top();
                        ss.pop();
                    }
                    else
                    {
                        sum += she - s.top();
                        she = s.top();
                        s.pop();

                    }
                }
            }
        }
        cout << "Case " << T << ": " << sum << endl;
    }
    return 0;
}

HDU 4302 貪吃蛇