POJ 3067 Japan 【树状数组经典】

题目链接：POJ 3067 Japan

Japan

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 32076		Accepted: 8620

Description

Japan plans to welcome the ACM ICPC World Finals and a lot of roads must be built for the venue. Japan is tall island with N cities on the East coast and M cities on the West coast (M <= 1000, N <= 1000). K superhighways will be build. Cities on each coast are numbered 1, 2, ... from North to South. Each superhighway is straight line and connects city on the East coast with city of the West coast. The funding for the construction is guaranteed by ACM. A major portion of the sum is determined by the number of crossings between superhighways. At most two superhighways cross at one location. Write a program that calculates the number of the crossings between superhighways.

Input

The input file starts with T - the number of test cases. Each test case starts with three numbers – N, M, K. Each of the next K lines contains two numbers – the numbers of cities connected by the superhighway. The first one is the number of the city on the East coast and second one is the number of the city of the West coast.

Output

For each test case write one line on the standard output:
Test case (case number): (number of crossings)

Sample Input

Sample Output

Test case 1: 5

Source

Southeastern Europe 2006

题目大意：

西岸有N座城市，东岸有M座城市，两岸之间有K条公路，问公路得总交点数。

大概思路：

假设公路连接的西岸城市为X，东岸城市为Y，则当 X_1 < X_2 && Y_2 < Y_1 时两公路相交.

那么问题到这里就差不多解决啦，先按 X 排一个升序，当 X 相等时按 Y 排升序。排序完成之后，求 Y 序列的逆序数即为交点总数。

注意事项：

①打代码时一定要心平气和

②树状数组是用来求Y的逆序数的所以范围是Y的范围，注意add（）函数的边界

③记得初始化

AC code（6656K 500ms）：

 1 #include <map>
 2 #include <set>
 3 #include <vector>
 4 #include <cstdio>
 5 #include <cstring>
 6 #include <iostream>
 7 #include <algorithm>
 8 #define ll long long int
 9 #define INF 0x3f3f3f3f
10 using namespace std;
11 
12 const int MAXN = 1002;
13 int N, M, K;
14 int c[MAXN*MAXN];
15 
16 struct node
17 {
18     int x, y;
19 }num[MAXN*MAXN];
20 
21 bool cmp(struct node a, struct node b)
22 {
23     if(a.x == b.x)
24         return a.y <= b.y;
25     return a.x < b.x;
26 }
27 
28 int lowbit(int x)
29 {
30     return x&(-x);
31 }
32 
33 void add(int i, int value)
34 {
35     while(i <= M)
36     {
37         c[i]+=value;
38         i+=lowbit(i);
39     }
40 }
41 
42 ll sum(int i)
43 {
44     ll res = 0;
45     while(i > 0)
46     {
47         res+=(ll)c[i];
48         i-=lowbit(i);
49     }
50     return res;
51 }
52 
53 int main()
54 {
55     int T_case = 0;
56     int t = 0;
57     scanf("%d", &T_case);
58     while(T_case--)
59     {
60         t++;
61         scanf("%d%d%d", &N, &M, &K);
62         memset(c, 0, sizeof(c));
63         for(int i = 1; i <= K; i++)
64         {
65             scanf("%d%d", &num[i].x, &num[i].y);
66         }
67         sort(num+1, num+K+1, cmp);
68         ll ans = 0;
69         for(int i = 1; i <= K; i++)
70         {
71             add(num[i].y, 1);
72             ans+=(i-sum(num[i].y));
73         }
74         printf("Test case %d: %lld\n", t, ans);
75     }
76     return 0;
77 }

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