poj1990两个树状数组

垃圾poj交不上去

/*
按权值从小到大排序，
两个树状数组维护权值小于等于并且在i左边的点的个数和权值
*/
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
#define maxn 20005
#define ll long long
struct node{
    int w,x;
    bool operator<(const node & a)const {
        return w<a.w;
    }
}a[maxn];
int n;
ll bitcnt[maxn],bittot[maxn],sum;
void add1(int x,int val){//bitcnt
    for(int i=x;i<=20001;i+=i&-i)
        bitcnt[i]+=val;
} 
void add2(int x,int val){//bittot
    for(int i=x;i<=20001;i+=i&-i)
        bittot[i]+=val;
}
ll query1(int x){
    ll res=0;
    for(int i=x;i;i-=i&-i)
        res+=bitcnt[i];
    return res;
}    
ll query2(int x){
    ll res=0;
    for(int i=x;i;i-=i&-i)
        res+=bittot[i];
    return res;
}    

int main(){
    while(scanf("%d",&n)==1){
        memset(bitcnt,0,sizeof bitcnt);
        memset(bittot,0,sizeof bittot);
        sum=0;
        
        for(int i=0;i<n;i++) scanf("%d%d",&a[i].w,&a[i].x);
        sort(a,a+n);
        
        ll ans=0;
        for(int i=0;i<n;i++){
            ll left_node=query1(a[i].x);
            ll left_total=query2(a[i].x);
            
            ans+=a[i].w*(left_node*a[i].x-left_total);
            ans+=a[i].w*((sum-left_total-(i-left_node)*a[i].x));
            sum+=a[i].x;
            add1(a[i].x,1);
            add2(a[i].x,a[i].x);
        }
        printf("%lld
",ans);
    }    
    return 0;
}