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题目大意：有n个数排成一列，问从中能找出几个三元组(ai,aj,ak)满足i<j<k并且这三个数严格单调。
题目分析：枚举中间的数字aj，如果aj前面有c(j)个数a(j)小，后面有d(j)个数比a(j)小，那么aj为中间数时，共有c(j)*(n-j-d(j))+d(j)*(j-1-c(j))。定义x(i)表示 i 出现过几次，当枚举到aj 时，只需统计sum(x(1)~x(a(j)-1)便得c(j)，d(j)求法类似。
代码如下：
# include<iostream>
# include<cstdio>
# include<cstring>
# include<vector>
# include<queue>
# include<list>
# include<set>
# include<map>
# include<string>
# include<cmath>
# include<cstdlib>
# include<algorithm>
using namespace std;
# define LL long long
const int N=1005;
const int INF=1000000000;
const LL oo=0x7fffffffffffffff;
const double eps=1e-10;
int n;
int maxn;
int a[N*100];
int b[N*100];
int c[N*100];
int d[N*100];
int lowbit(int x)
{
return (x&(-x));
}
int Query(int x)
{
int res=0;
while(x>0)
{
res+=b[x];
x-=lowbit(x);
}
return res;
}
void Update(int x,int k)
{
while(x<=maxn)
{
b[x]+=k;
x+=lowbit(x);
}
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
maxn=0;
for(int i=1;i<=n;++i){
scanf("%d",a+i);
maxn=max(maxn,a[i]);
}
memset(b,0,sizeof(b));
for(int i=1;i<=n;++i){
c[i]=Query(a[i]-1);
Update(a[i],1);
}
memset(b,0,sizeof(b));
for(int i=n;i>=1;--i){
d[i]=Query(a[i]-1);
Update(a[i],1);
}
LL ans=0;
for(int i=1;i<=n;++i){
ans+=c[i]*(n-i-d[i]);
ans+=d[i]*(i-1-c[i]);
}
printf("%lld\n",ans);
}
return 0;
}
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原文地址：http://www.cnblogs.com/20143605--pcx/p/5432844.html
