POJ-3928 Ping pong(树状数组)

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <queue>
 6 #include <stack>
 7 #include <vector>
 8 #include <iostream>
 9 #include "algorithm"
10 using namespace std;
11 typedef long long LL;
12 const int MAX=100005;
13 int T;
14 int n;
15 int c[MAX];
16 struct Peo{
17     int name;
18     int sco;
19     bool operator <(const Peo &zt) const {
20         if (sco!=zt.sco)
21          return sco<zt.sco;
22         return name<zt.name;
23     }
24 }stu[MAX];
25 void add(int x,int y){
26     for (;x<=n;c[x]+=y,x+=(x&-x));
27 }
28 LL search(int x){
29     LL an(0);
30     for (;x>0;an+=c[x],x-=(x&-x));
31     return an;
32 }
33 int main(){
34     freopen ("tennis.in","r",stdin);
35     freopen ("tennis.out","w",stdout);
36     int i,j;
37     scanf("%d",&T);
38     while (T--){
39     scanf("%d",&n);
40     for (i=1;i<=n;i++)
41     {scanf("%d",&stu[i].sco);
42      stu[i].name=i;
43     }
44     sort(stu+1,stu+n+1);
45     memset(c,0,sizeof(c));
46     LL x1,x2,x3,x4,ans(0);
47     add(stu[1].name,1);
48     for (i=2;i<=n;i++)
49     {x1=search(stu[i].name-1);
50      x2=stu[i].name-1-x1;
51      x3=search(n)-search(stu[i].name);
52      x4=n-stu[i].name-x3;
53      ans=ans+x1*x4+x2*x3;
54      add(stu[i].name,1);
55     }
56     printf("%lld\n",ans);
57     }
58     return 0;
59 }