agc012d

题目
思路：
首先交换是可传递的，所以我们将所有和满足条件的点对连起来就是对的，复杂度\(O(n^2)\)。
考虑优化建边：最小颜色的最小球往其他颜色的最小球连，相同颜色往最小连，最小颜色非最小球往次小颜色最小颜色连。

难点主要在优化

#include<bits/stdc++.h>
using namespace std;
struct FIO{static const int BUF_SIZE=1<<19;char _i[BUF_SIZE],*_1=_i,*_2=_i,_o[BUF_SIZE],*_t=_o;inline char _gt(){if(_1==_2){_1=_i;_2=_i+fread(_i,1,BUF_SIZE,stdin);}return _1==_2?EOF:*_1++;}inline void _pt(char c){if(_t-_o==BUF_SIZE){fwrite(_o,1,BUF_SIZE,stdout);_t=_o;}*_t++=c;}int _pr=6,le,_ob[20];template<typename T>FIO&operator>>(T&x){x=0;char c=_gt();bool sg=false;while(c!=EOF&&(c<'0'||c>'9')){if(c=='-')sg=true;c=_gt();}while(c!=EOF&&c>='0'&&c<='9'){x=(x<<1)+(x<<3)+(c^48);c=_gt();}if(sg)x=-x;return*this;}FIO&operator>>(char*s){char c=_gt();while(c!=EOF&&c<33)c=_gt();while(c!=EOF&&c>=33)*s++=c,c=_gt();*s='\0';return*this;}FIO&operator>>(string&s){s.clear();char c=_gt();while(c!=EOF&&c<33)c=_gt();while(c!=EOF&&c>=33)s+=c,c=_gt();return*this;}FIO&operator>>(char&x){char c=_gt();while(c!=EOF&&c<33)c=_gt();x=c;return*this;}FIO&operator>>(double&x){x=0;char c=_gt();bool sg=0;while(c!=EOF&&c<33)c=_gt();if(c=='-'){sg=1;c=_gt();}while(c!=EOF&&c>='0'&&c<='9'){x=x*10+(c^48);c=_gt();}if(c=='.'){c=_gt();double tp=0.1;while(c!=EOF&&c>='0'&&c<='9'){x+=(c^48)*tp;tp*=0.1;c=_gt();}}if(sg)x=-x;return*this;}FIO&setprecision(int n){_pr=n<0?0:n;return*this;}template<typename T>FIO&operator<<(T x){if(x<0){_pt('-');x=-x;}if(x==0){_pt('0');return*this;}char _b[20];int ps=0;while(x>0){_b[ps++]=(x%10)^48;x/=10;}while(ps>0)_pt(_b[--ps]);return*this;}FIO&operator<<(const char*s){while(*s)_pt(*s++);return*this;}FIO&operator<<(const string&s){for(char c:s)_pt(c);return*this;}FIO&operator<<(char c){_pt(c);return*this;}FIO&operator<<(double x){if(x<0)_pt('-'),x=-x;if(isnan(x))return _pt('n'),_pt('a'),_pt('n'),*this;if(isinf(x))return _pt('i'),_pt('n'),_pt('f'),*this;long long ip=(long long)x;x-=ip;if(_pr==0){x*=10;if(x>=5)ip++;return*this<<ip;}le=0;_ob[0]=0;for(int i=1;i<=_pr;i++){x*=10;_ob[++le]=(int)x;x-=(int)x;}x*=10;int _tp=(int)x;if(_tp>=5){int j=le;_ob[le]++;while(_ob[j]==10)_ob[j-1]++,_ob[j]=0,j--;}ip+=_ob[0];*this<<ip<<".";for(int i=1;i<=le;i++)*this<<_ob[i];return*this;}~FIO(){fwrite(_o,1,_t-_o,stdout);}}fst;
#define cin fst
#define cout fst
#define il inline
#define mod 1000000007
#define inf (INT_MAX>>1)
typedef long long ll;
ll qpow(ll x,ll e){
	ll res=1;
	while(e){
		if(e&1)res=res*x%mod;
		x=x*x%mod;
		e>>=1;
	}
	return res;
}
#define N 1000005
int n,x,y;
ll fac[N],invf[N];
il void init(int n){
	fac[0]=1;
	for(int i=1;i<=n;i++)fac[i]=fac[i-1]*i%mod;
	invf[n]=qpow(fac[n],mod-2);
	for(int i=n-1;i>=0;i--)invf[i]=invf[i+1]*(i+1)%mod;
}
int mn[N];
vector<int> bs[N];
int main(){
	cin>>n>>x>>y;
	init(n);
	for(int i=1;i<=n;i++)mn[i]=inf;
	for(int i=1,c,w;i<=n;i++){
		cin>>c>>w;
		mn[c]=min(mn[c],w);
		bs[c].push_back(w);
	}
	int mni=0,nxi=0;
	mn[0]=inf;
	for(int i=1;i<=n;i++){
		if(mn[i]<mn[mni])nxi=mni,mni=i;
		else if(mn[i]<mn[nxi])nxi=i;
	}
	ll c1=0,c2=1;
	for(int i=1;i<=n;i++){
		if(mn[i]+mn[mni]>y)continue;
		int cnt=0;
		for(int v:bs[i]){
			cnt+=(v+mn[mni==i?nxi:mni]<=y||v+mn[i]<=x);
		}
		c1+=cnt,c2=c2*invf[cnt]%mod;
	}
	cout<<fac[c1]*c2%mod;
	return 0;
}