三元逆序对求ia[j]>a[k] 的对数树状数组Codeforces 61E Enemy is weak

71 阅读 0 评论 47 点赞

我是靠谱客的博主怕孤单狗，最近开发中收集的这篇文章主要介绍三元逆序对求ia[j]>a[k] 的对数树状数组Codeforces 61E Enemy is weak，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

http://codeforces.com/problemset/problem/61/E

The Romans have attacked again. This time they are much more than the Persians but Shapur is ready to defeat them. He says: "A lion is never afraid of a hundred sheep".

Nevertheless Shapur has to find weaknesses in the Roman army to defeat them. So he gives the army a weakness number.

In Shapur's opinion the weakness of an army is equal to the number of triplets i, j, k such that i < j < k and ai > aj > ak where ax is the power of man standing at position x. The Roman army has one special trait — powers of all the people in it are distinct.

Help Shapur find out how weak the Romans are.

Input

The first line of input contains a single number n (3 ≤ n ≤ 106) — the number of men in Roman army. Next line contains n different positive integers ai (1 ≤ i ≤ n, 1 ≤ ai ≤ 109) — powers of men in the Roman army.

Output

A single integer number, the weakness of the Roman army.

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).

Sample test(s)

input
3
3 2 1

output
1

input
3
2 3 1

output
0

input
4
10 8 3 1

output
4

input
4
1 5 4 3

output
1

题意是求 i<j<k && a[i]>a[j]>a[k] 的对数

如果只有2元组那就是求逆序数的做法

三元组的话就用一个树状数组x表示数字i前面有多少个比自己大的个数

然后每次给这个y数组求和，再把x中>a[i]的个数存入y中即可

即使是4元也可以求了

#include <algorithm>
#include <cctype>
#include <cassert>
#include <cstdio>
#include <cstring>
#include <climits>
#include <vector>
#include<iostream>
using namespace std;
#define ll long long
inline void rd(int &ret)
{
char c;
do { c = getchar();
} while(c < '0' || c > '9');
ret = c - '0';
while((c=getchar()) >= '0' && c <= '9')
ret = ret * 10 + ( c - '0' );
}
#define N 1000005
#define eps 1e-8
#define inf 1000000
ll n;
struct node{
ll c[N];
inline ll lowbit(ll x){return x&-x;}
void init(){memset(c, 0, sizeof c);}
ll sum(ll x){
ll ans = 0;
while(x<=n+10)
ans += c[x], x+=lowbit(x);
return ans;
}
void change(ll x, ll y){
while(x)
c[x] +=y, x-=lowbit(x);
}
}x, y;
int haifei[1000000], panting[1000000];
int main()
{
ll i, j;
while(cin>>n)
{
ll ans = 0;
for(i = 0; i < n; i++)rd(haifei[i]), panting[i] = haifei[i];
x.init(); y.init();
sort(haifei, haifei+n);
for(i = 0; i < n; i++)
{
ll b = (lower_bound(haifei, haifei+n, panting[i]) - haifei) +1;
ll siz = y.sum(b);
ans += siz;
y.change(b, x.sum(b));
x.change(b, 1);
}
cout<<ans<<endl;
}
return 0;
}