POJ 3067 Japan 【 树状数组 】

题意：左边有n个城市，右边有m个城市，现在修k条路，问会形成多少个交点

先按照x从小到大排，x相同的话，则按照y从小到大排，然后对于每一个y统计前面有多少个y比它大，它们就一定会相交

另外要用long long

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include <cmath>
 5 #include<stack>
 6 #include<vector>
 7 #include<map>
 8 #include<set>
 9 #include<queue>
10 #include<algorithm>
11 using namespace std;
12
13 typedef long long LL;
14 const int INF = (1<<30)-1;
15 const int mod=1000000007;
16 const int maxn=1000005;
17
18 int a[maxn];
19 int c[maxn];//树状数组
20 int n,m,k;
21
22 struct node{
23     int x,y;
24 }p[maxn];
25
26 int cmp(node n1,node n2){
27     if(n1.x != n2.x) return n1.x < n2.x;
28     return n1.y < n2.y;
29 }
30
31 int lowbit(int x){ return x & (-x);}
32
33 int sum(int x){
34     int ret =0;
35     while(x > 0){
36         ret+=c[x];x-=lowbit(x);
37     }
38     return ret;
39 }
40
41 void add(int x,int d){
42     while(x < maxn){
43         c[x]+=d;x+=lowbit(x);
44     }
45 }
46
47 int main(){
48     int T;
49     scanf("%d",&T);
50     int kase=0;
51     while(T--){
52         scanf("%d %d %d",&n,&m,&k);
53         for(int i=1;i<=k;i++) scanf("%d %d",&p[i].x,&p[i].y);
54         sort(p+1,p+k+1,cmp);
55
56         memset(c,0,sizeof(c));
57         LL ans=0;
58         for(int i=1;i<=k;i++){
59             add(p[i].y,1);
60             ans += i- sum(p[i].y);
61         }
62         printf("Test case %d: %I64d\n",++kase,ans);
63     }
64     return 0;
65 }