poj 3067 Japan（树状数组）

 1 /*************************************************************************
 2     > File Name: code/poj/3067.cpp
 3     > Author: 111qqz
 4     > Email: rkz2013@126.com 
 5     > Created Time: 2015年08月04日 星期二 10时49分06秒
 6  ************************************************************************/
 7 
 8 #include<iostream>
 9 #include<iomanip>
10 #include<cstdio>
11 #include<algorithm>
12 #include<cmath>
13 #include<cstring>
14 #include<string>
15 #include<map>
16 #include<set>
17 #include<queue>
18 #include<vector>
19 #include<stack>
20 #define y0 abc111qqz
21 #define y1 hust111qqz
22 #define yn hez111qqz
23 #define j1 cute111qqz
24 #define tm crazy111qqz
25 #define lr dying111qqz
26 using namespace std;
27 #define REP(i, n) for (int i=0;i<int(n);++i)  
28 typedef long long LL;
29 typedef unsigned long long ULL;
30 const int inf = 0x7fffffff;
31 const int N=1E3+5;
32 
33 int n,m,k;
34 int t[N*N/2];
35 struct R
36 {
37     int x,y;
38 }r[N*N/2];
39 
40 bool cmp(R a,R b)
41 {
42     if (a.x<b.x) return true;
43     if (a.x==b.x&&a.y<b.y) return true;
44     return false;
45 }
46 int lowbit(int x)
47 {
48     return x&(-x);
49 }
50 void update (int x,int c)
51 {
52     for ( int i = x ; i < N ; i = i + lowbit(i))
53     {
54     t[i] = t[i] + c;
55     }
56 }
57 LL sum ( int x)
58 {
59     int res = 0 ;
60     for ( int i = x ; i >= 1 ; i = i - lowbit(i))  
61     {
62     res = res + t[i];
63     }
64     return res;
65 }
66 int main()
67 {
68     int T;
69     cin>>T;
70     LL ans ;
71     int cas = 0;
72     while (T--)
73     {
74     memset(t,0,sizeof(t));
75     memset(r,0,sizeof(r));
76     ans =  0 ;
77     cas++;
78     scanf("%d %d %d",&n,&m,&k);
79     for ( int i = 0 ; i  < k ; i ++ )
80     {
81         scanf("%d %d",&r[i].x,&r[i].y);
82     }
83     sort(r,r+k,cmp);
84     for ( int i = 0 ; i < k;  i ++ )  //起点终点相同不算crossing
85     {
86       //  update(r[i].x,1);
87        // if (r[i].x==r[i-1].x||r[i].y==r[i-1].y) continue;
88         ans = ans +i- sum(r[i].y);  //求y的逆序对数，好好理解这行代码！
89     //    cout<<"ans:"<<ans<<endl;
90         update(r[i].y,1);
91     }
92     printf("Test case %d: %lld\n",cas,ans);
93 
94     }
95     return 0;
96 }