#树状数组#poj 3928 ping pong

题目

对于$1\leq x<y<z\leq n$,当$a_x>a_y>a_z或a_x<a_y<a_z$，累计方案，求总方案。

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分析

可以用树状数组维护$a_i左右边的最小值和最大值$，

left_min	left_max	right_min	right_max
b[i]	i-1-b[i]	d[i]	n-i-d[i]

跑O（n）算出$(i-1-b[i])*d[i]+b[i]*(n-i-d[i])$,时间复杂度O（nlogn）

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代码

#include <cstdio>
#include <cctype>
#include <cstring>
using namespace std;
int a[20001],b[20001],c[100001],d[20001],n,t;
int in(){
    int ans=0; char c=getchar();
    while (!isdigit(c)) c=getchar();
    while (isdigit(c)) ans=ans*10+c-48,c=getchar();
    return ans;
}
void add(int x){while (x<=100000) c[x]++,x+=-x&x;}
int answer(int x){int ans=0; while (x) ans+=c[x],x-=-x&x; return ans;}
int main(){
    t=in(); unsigned long long ans;
    while (t--){
        n=in(); memset(c,0,sizeof(c)); ans=0; 
        for (int i=1;i<=n;i++) a[i]=in(),b[i]=answer(a[i]-1),add(a[i]);
        for (int i=1;i<=n;i++) d[i]=answer(a[i]-1)-b[i];
        for (int i=1;i<=n;i++) ans+=(i-1-b[i])*d[i]+b[i]*(n-i-d[i]);
        printf("%lld\n",ans);
    }
    return 0;
}