HDU5792 World is Exploding(树状数组)

     一共6种情况，a < b且Aa < Ab, c < d 且Ac > Ad，这两种情况数量相乘，再减去a = c, a = d, b = c, b = d这四种情况，使用树状数组维护，le[i]表示i左边比他小的数数量，le1[i]表示i左边比他大的数数量，ri[i]表示i右边比他小的数数量，ri1[i]表示i右边比他大的数数量。

　　跑两次树状数组，求出这四个数组值。

　　注意处理数值相等的情况。

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<utility>
using namespace std;
typedef long long LL;
const int N = 60008, INF = 0x3F3F3F3F;
#define MS(a, num) memset(a, num, sizeof(a))
#define PB(A) push_back(A)
#define FOR(i, n) for(int i = 0; i < n; i++)
int C[N];
int n;
int le[N], ri[N], le1[N], ri1[N];
int val[N];
int b[N], tp[N];
inline int lowbit(int x){
    return x&-x;
}
inline void add(int x, int val){
    for(int i=x;i<=n;i+=lowbit(i)){
        C[i] += val;
    }
}
inline int sum(int x){
    int ret = 0;
    for(int i=x;i>0;i-=lowbit(i)){
        ret+=C[i];
    }
    return ret;
}

int main(){
    while(~scanf("%d", &n)){
        for(int i = 1; i <= n; i++){
            scanf("%d", &val[i]);
            tp[i] = val[i];
        }

        sort(val + 1, val + n + 1);
        for(int i = 1; i <= n; i++){
            b[i]  = lower_bound(val + 1, val + n + 1, tp[i]) - val;
        }
        MS(C , 0);
        for(int i = 1; i <= n; i++){
            le[i] = sum(b[i] - 1);
            le1[i] = sum(n) - sum(b[i]);
            add(b[i], 1);
        }

        MS(C, 0);
        for(int i = n; i >= 1; i--){
            ri[i] = sum(b[i] - 1);
            ri1[i] = sum(n) - sum(b[i]);
            add(b[i], 1);
        }
        LL ans= 0;
        LL sum = 0;
        for(int i = n; i >= 1; i--){
            sum += ri1[i];
        }
        ans = sum;
        sum = 0;
        for(int i = n; i >= 1; i--){
            sum += ri[i];
        }
        ans *= sum;
        sum = 0;
        for(int i = 1; i <= n; i++){
            sum += (LL)ri[i] * (LL)ri1[i];
            sum += (LL)ri1[i] * (LL)le1[i];
            sum += (LL)le[i] * (LL)le1[i];
            sum += (LL)le[i] * (LL)ri[i];
        }
        ans -= sum;
        printf("%I64d\n", ans);
    }
    return 0;
}