SGU 180 Inversions【树状数组】

题意：求逆序数

和POJ那题求逆序数的一样，不过这题离散化之后，要去一下重，然后要开到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=80005;

17 

18 int a[maxn];

19 int c[maxn];//树状数组

20 

21 struct node{

22     int x;

23     int id;

24 } p[maxn];

25 

26 int cmp(node n1,node n2){

27     return n1.x < n2.x;

28 }

29 

30 int lowbit(int x){ return x &(-x);}

31 

32 __int64 sum(int x){

33     __int64 ret=0;

34     while(x > 0){

35         ret+=c[x];x-=lowbit(x);

36     }

37     return ret;

38 }

39 

40 void add(int x,int d){

41     while(x < maxn){

42         c[x]+=d;x+=lowbit(x);

43     }

44 }

45 

46 int main(){

47     int n;

48     scanf("%d",&n);

49     for(int i=1;i<=n;i++) scanf("%d",&p[i].x),p[i].id=i;

50     sort(p+1,p+n+1,cmp);

51     

52     for(int i=1,j=0;i<=n;i++){

53         if(i==1 || p[i].x != p[i-1].x) j++;

54         a[p[i].id] = j;

55     }

56     

57     memset(c,0,sizeof(c));

58     __int64 ans=0;

59     for(int i=1;i <=n;i++ ) {

60         add(a[i],1);

61         ans += i - sum(a[i]); 

62     }

63     printf("%I64d\n",ans);

64     return 0;

65 } 

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