HDU 4000 Fruit Ninja 树状数组

求逆序对的加强版.

先求出 x<y<z和x<z<y的总数，再求出x<y<z的总数，相减得答案。

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define pb(a) push_back(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c) {
    return min(min(a,b),min(a,c));
}
template<class T> T max(const T& a,const T& b,const T& c) {
    return max(max(a,b),max(a,c));
}
void debug() {
#ifdef ONLINE_JUDGE
#else

    freopen("d:\\in.txt","r",stdin);
   // freopen("d:\\out1.txt","w",stdout);
#endif
}
int getch() {
    int ch;
    while((ch=getchar())!=EOF) {
        if(ch!=' '&&ch!='\n')return ch;
    }
    return EOF;
}

const int maxn=100100;
const int mod=100000007;
int x[maxn];
int da[maxn];
int c[maxn];
int d[maxn];
int n;
int add(int a)
{
    while(a<=n)
    {
        x[a]++;
        a=a+(a&-a);
    }
}
int sum(int a)
{
    int res=0;
    while(a>0)
    {
        res+=x[a];
        a=a-(a&-a);
    }
    return res;
}
int main()
{
    int t;
    scanf("%d",&t);
    for(int ca=1;ca<=t;ca++)
    {
        scanf("%d",&n);
        int res=0;
        for(int i=1;i<=n;i++)
            scanf("%d",&da[i]);
        memset(x,0,sizeof(x));
        for(int i=n;i>=1;i--)
        {
            ll a=n-i-sum(da[i]-1);
            res=((ll)res+a*(a-1)/2)%mod;
            add(da[i]);
        }
        memset(x,0,sizeof(x));
        for(int i=1;i<=n;i++)
        {
            c[i]=sum(da[i]-1);
            add(da[i]);
        }
        memset(x,0,sizeof(x));
        for(int i=n;i>=1;i--)
        {
            d[i]=n-i-sum(da[i]);
            add(da[i]);
        }
        for(int i=1;i<=n;i++)
        {
            res=(res+mod-c[i]*d[i])%mod;
        }
        printf("Case #%d: %d\n",ca,res);
    }
    return 0;
}