CodeForces - 283E Cow Tennis Tournament

Discription

Farmer John is hosting a tennis tournament with his n cows. Each cow has a skill level s_i, and no two cows having the same skill level. Every cow plays every other cow exactly once in the tournament, and each cow beats every cow with skill level lower than its own.

However, Farmer John thinks the tournament will be demoralizing for the weakest cows who lose most or all of their matches, so he wants to flip some of the results. In particular, at k different instances, he will take two integers a_i,?b_i (a_i?<?b_i) and flip all the results between cows with skill level between a_i and b_i inclusive. That is, for any pair x,?y  he will change the result of the match on the final scoreboard (so if x won the match, the scoreboard will now display that ywon the match, and vice versa). It is possible that Farmer John will change the result of a match multiple times. It is not guaranteed that a_i and b_i are equal to some cow‘s skill level.

Farmer John wants to determine how balanced he made the tournament results look. In particular, he wants to count the number of triples of cows (p,?q,?r) for which the final leaderboard shows that cow p beats cow q, cow q beats cow r, and cow r beats cow p. Help him determine this number.

Note that two triples are considered different if they do not contain the same set of cows (i.e. if there is a cow in one triple that is not in the other).

Input

On the first line are two space-separated integers, n and k (3?≤?n?≤?10⁵; 0?≤?k?≤?10⁵). On the next line are n space-separated distinct integers, s₁,?s₂,?...,?s_n (1?≤?s_i?≤?10⁹), denoting the skill levels of the cows. On the next k lines are two space separated integers, a_i and b_i (1?≤?a_i?<?b_i?≤?10⁹) representing the changes Farmer John made to the scoreboard in the order he makes it.

Output

A single integer, containing the number of triples of cows (p,?q,?r) for which the final leaderboard shows that cow p beats cow q, cow q beats cow r, and cow r beats cow p.

Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

3 21 2 31 22 3

Output

1

Input

5 35 9 4 1 71 72 83 9

Output

3

Note

In the first sample, cow 3 > cow 1, cow 3 > cow 2, and cow 2 > cow 1. However, the results between cows 1 and 2 and cows 2 and 3 are flipped, so now FJ‘s results show that cow 1 > cow 2, cow 2 > cow 3, and cow 3 > cow 1, so cows 1, 2, and 3 form a balanced triple.

jzh大佬给学弟学妹们讲课的课件里，唯一一个不是弱智题的就是这个了2333，然鹅一找原题，mdzz数据范围后面加了俩0，有毒。。。

如果n<=1000的话，我们可以很容易的用差分去维护区间覆盖的问题，然后暴力计算每两个牛之间的比赛结果就好了。。。

所以n<=1e5怎么做呢？？

我们只要先求出每个人最后赢的场数，就可以直接算出不合法的三元组数量，再用C(n,3)减去这个就是答案了。

那么如何快速计算每个人赢的场数呢？

考虑扫描线，把修改存在vector里，先倒着扫一遍，查询s[j] < s[i] 且 i赢j的个数；再倒着扫一遍，。。。。把修改看成区间异或，查询看成区间1的个数，然后这就是基本线段树操作了23333

#include<bits/stdc++.h>
#define ll long long
using namespace std;
#define pb push_back
#define lc (o<<1)
#define mid (l+r>>1)
#define rc ((o<<1)|1)
const int maxn=100005;
vector<int> L[maxn],R[maxn];
int a[maxn],n,num[maxn],k,X,Y,len[maxn*4];
int win[maxn],le,ri,sum[maxn*4],tag[maxn*4],w;
ll ans=0;

inline void maintain(int o){ sum[o]=sum[lc]+sum[rc];}

inline void work(int o){ tag[o]^=1,sum[o]=len[o]-sum[o];}

inline void pushdown(int o){
	if(tag[o]){
		tag[o]=0;
		work(lc),work(rc);
	}
}

void build(int o,int l,int r){
	len[o]=r-l+1;
	if(l==r){ sum[o]=1; return;}
	build(lc,l,mid);
	build(rc,mid+1,r);
	maintain(o);
}

void update(int o,int l,int r){
	if(l>=le&&r<=ri){ work(o); return;}
	pushdown(o);
	if(le<=mid) update(lc,l,mid);
	if(ri>mid) update(rc,mid+1,r);
	maintain(o);
}

void query(int o,int l,int r){
	if(l>=le&&r<=ri){ w+=sum[o]; return;}
	pushdown(o);
	if(le<=mid) query(lc,l,mid);
	if(ri>mid) query(rc,mid+1,r);
}

inline void solve(){
	build(1,1,n);

	for(int i=n;i;i--){
		ri=i;

		for(int j=L[i].size()-1;j>=0;j--) le=L[i][j],update(1,1,n);

		w=0,le=1,ri=i-1;
		if(le<=ri) query(1,1,n);
		win[i]+=w;

//		cout<<i<<‘ ‘<<w<<endl;
	}

	memset(sum,0,sizeof(sum));
	memset(tag,0,sizeof(tag));

	for(int i=1;i<=n;i++){
		le=i;
		for(int j=R[i].size()-1;j>=0;j--) ri=R[i][j],update(1,1,n);

		w=0,le=i+1,ri=n;
		if(le<=ri) query(1,1,n);
		win[i]+=w;

//		cout<<i<<‘ ‘<<win[i]<<endl;
	}

	for(int i=1;i<=n;i++) ans-=win[i]*(ll)(win[i]-1)>>1;
}

int main(){
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++) scanf("%d",a+i),num[i]=a[i];

	sort(num+1,num+n+1);
//	unique(num+1,num+n+1);
	for(int i=1;i<=n;i++) a[i]=lower_bound(num+1,num+n+1,a[i])-num;

	while(k--){
		scanf("%d%d",&X,&Y);
		X=lower_bound(num+1,num+n+1,X)-num;
		Y=upper_bound(num+1,num+n+1,Y)-num-1;

		if(!X||!Y) continue;

		L[Y].pb(X),R[X].pb(Y);
	}

	ans=n*(ll)(n-1)*(ll)(n-2)/6ll,solve();

	cout<<ans<<endl;

	return 0;
}

原文地址：https://www.cnblogs.com/JYYHH/p/9096479.html