题目：
| Collisions |
| Time Limit: 10000ms, Special Time Limit:25000ms, Memory Limit:65536KB |
| Total submit users: 10, Accepted users: 7 |
| Problem 12264 : No special judgement |
| Problem description |
| Identical small balls are located on a straight line and can move along this line only. Each ball moves with a constant velocity, but velocities of different balls may be different. When two balls meet, a perfectly elastic collision occurs. It’s a common-known physical fact, that when two equal-mass physical bodies A and B collide perfectly elastically, they swap their velocities, i.?e. new A’s velocity is old B’s one, and new B’s is old A’s. Your task is to write a program to find the total number of collisions. |
| Input |
| The first line at input contains the number of balls N (3 ≤ N ≤ 2000000). Each of the following N lines contains 2 space-separated integers — the starting coordinate and the velocity of corresponding ball. All start coordinates are in range -1011 < xi < 1011, all velocities are in range -108 < vi < 108. All start coordinates are different. It’s guaranteed that each collision involves exactly two balls (none involves three or more balls together). |
| Output |
| Your program should output exactly one integer number in a single line ? the total number of collisions (or 987654321987654321 if the number is infinite). |
| Sample Input |
3 -5 3 0 -1 7 -2 |
| Sample Output |
3 |
| Problem Source |
|
2010 All-Ukrainian Collegiate Programming Contest Vinnytsia National Technical University |
题意：就是告诉你有n，个小球，第二行是n个小球的位置和速度，所有小球都在一条直线上，小球相碰后交换速度，问最多有几次碰撞。如果无数次，输出987654321987654321.
思路：开始毫无思路，后来想了一下，相碰后交换速度，不是等价于直接穿过去吗，如果把小球按位置排序，即从小到大排序，然后求速度的逆序数不就行了吗？比如两个小球，一前一后分别为5,4，者后面小球会追上相碰，或者4，-3，也一样，还有，-1，-2，也是那个道理，所以题目就是叫你求逆序数。那样就不会有无数次的情况了，真的吗？答案是对的。求逆序数有两种方法，归并排序和树状数组，至少我知道这两种，不太会用，归并用了个模板，一次就过了，还是挺惊讶的。
代码：
1 #include<iostream>
2 #include<algorithm>
3 using namespace std;
4 const int size=2000500;
5 __int64 date[size];
6 struct point{
7 __int64 position,v;
8 };
9 point temp[size];
10 int n;
11 __int64 L[size];
12 __int64 R[size];
13 const int Max = 1 <<30;
14 __int64 change = 0;
15 bool cmp(point x,point y)
16 {
17 return x.position<y.position;
18 }
19
20 void Merge(__int64 *data,int left,int divide,int right)
21 {
22 int lengthL = divide - left;
23 int lengthR = right - divide;
24
25 for(int i = 0; i < lengthL; ++i)
26 {
27 L[i] = data[left + i];
28 }
29 for(int i = 0; i < lengthR; ++i)
30 {
31 R[i] = data[divide + i];
32 }
33 L[lengthL] = R[lengthR] = Max;
34 int i = 0;
35 int j = 0;
36 for(int k = left; k < right; ++k)
37 {
38 if(L[i] <= R[j])
39 {
40 data[k] = L[i];
41 ++i;
42 }
43 else
44 {
45 change += divide - i - left ;
46 data[k] = R[j];
47 ++j;
48 }
49 }
50
51 }
52
53 void MergeSort(__int64 *data,int left,int right)//归并排序求逆序数
54 {
55 if(left < right -1)
56 {
57 int divide = (left + right)/2;
58 MergeSort(data,left,divide);
59 MergeSort(data,divide,right);
60 Merge(data,left,divide,right);
61 }
62 }
63
64
65
66 int main()
67 {
68 scanf("%d",&n);
69 for(int i=0;i<n;i++)
70 scanf("%I64d%I64d",&temp[i].position,&temp[i].v);
71 sort(temp,temp+n,cmp);//按位置排序
72 for(int i=0;i<n;i++)
73 date[i]=temp[i].v;
74 MergeSort(date,0,n);
75 cout<<change<<endl;
76 //system("pause");
77 return 0;
78 }
***************************************************************************************
总结：
1.基本代码要会敲，比如归并
2.相信自己，比如不会有无数次的情况
3.多练
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