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 } 

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