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.
InputThe 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.
OutputOutput 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
题目意思:
有一条蛇在一条管道中,管道中有着一些果子,蛇每次吃果子都会选择距离最短的果子,如果有距离相同的就按前进方向的吃
解法:
我使用了两个优先队列来分别储存出现在蛇前面和后面的果子
1 #include <iostream>
2 #include <string.h>
3 #include <set>
4 #include <stack>
5 #include <queue>
6 using namespace std;
7
8 const int MAX = 5*100000 + 2000;
9 struct cmp
10 {
11 bool operator () (int a,int b)
12 {
13 return a>b;
14 }
15 };
16
17
18
19 int main()
20 {
21 int N;
22 cin>>N;
23 int N0 = 0;
24 while(N--)
25 {
26 int x = 0;
27 int L,n;
28 int ti = 1;
29 int sum =0;
30 priority_queue<int,vector<int>,cmp> p2;
31 priority_queue<int>p1;
32
33
34
35 cin>>L>>n;
36 while(n--)
37 {
38 int temp;
39 cin>>temp;
40 if(temp == 0)
41 {
42 cin>>temp;
43 if(temp >= x) { p2.push(temp); }
44 else { p1.push(temp); }
45 }
46 else if(temp == 1)
47 {
48 if(!p1.empty()&&!p2.empty())
49 {
50 if(x - p1.top() < p2.top() - x )
51 {
52 ti = -1;
53 sum += x - p1.top();
54 x = p1.top();
55 p1.pop();
56 }
57 else if( x - p1.top() > p2.top() - x)
58 {
59 ti = 1;
60 sum +=p2.top() - x;
61 x = p2.top();
62 p2.pop();
63
64 }
65 else if(x - p1.top() == p2.top() - x)
66 {
67 if(ti == 1)
68 {
69 ti = 1;
70 sum +=p2.top() - x;
71 x = p2.top();
72 p2.pop();
73 }
74 else if( ti == -1)
75 {
76 ti = -1;
77 sum += x - p1.top();
78 x = p1.top();
79 p1.pop();
80 }
81
82 }
83
84 }
85 else if(!p1.empty())
86 {
87 ti = -1;
88 sum += x - p1.top();
89 x = p1.top();
90 p1.pop();
91
92 }
93 else if(!p2.empty())
94 {
95 ti = 1;
96 sum +=p2.top() - x;
97 x = p2.top();
98 p2.pop();
99 }
100 }
101
102
103 // cout<<"x= sum ="<<" "<<x<<" "<<sum<<endl;
104 }
105 cout<<"Case "<<++N0<<": "<<sum<<endl;
106 }
107
108
109 return 0;
110 }