CodeForces 61E Enemy is weak 求i<j<k && a[i]>a[j]>a[k] 的对数 树状数组

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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

#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;
}

CodeForces 61E Enemy is weak 求i<j<k && a[i]>a[j]>a[k] 的对数 树状数组,布布扣,bubuko.com

CodeForces 61E Enemy is weak 求i<j<k && a[i]>a[j]>a[k] 的对数 树状数组

标签：blog   http   os   io   for   2014   ar   line   

原文地址：http://blog.csdn.net/qq574857122/article/details/38427881