結果

問題 No.329 全射
コンテスト
ユーザー shimomire
提出日時 2015-12-20 01:56:52
言語 C++11（廃止可能性あり）
(gcc 13.3.0 + boost 1.89.0)
結果
AC  
実行時間 219 ms / 2,000 ms
コード長 6,150 bytes
記録
記録タグの例:
初AC ショートコード 純ショートコード 純主流ショートコード 最速実行時間
コンパイル時間 1,292 ms
コンパイル使用メモリ 134,020 KB
実行使用メモリ 12,800 KB
最終ジャッジ日時 2024-09-17 11:17:19
合計ジャッジ時間 6,123 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 40
権限があれば一括ダウンロードができます

ソースコード

diff #
raw source code 

#include <cassert>// c
#include <iostream>// io
#include <iomanip>
#include <fstream>
#include <sstream>
#include <vector>// container
#include <map>
#include <set>
#include <queue>
#include <bitset>
#include <stack>
#include <algorithm>// other
#include <complex>
#include <numeric>
#include <functional>
#include <random>
#include <regex>
using namespace std;

typedef long long ll;typedef unsigned long long ull;typedef long double ld;

#define ALL(c) c.begin(),c.end()
#define IN(l,v,r) (l<=v && v < r)
template<class T> void UNIQUE(T& v){v.erase(unique(ALL(v)),v.end());}
//debug
#define DUMP(x) cerr << #x <<" = " << (x)
#define LINE() cerr<< " (L" << __LINE__ << ")"

struct range{
	struct Iter{
		int v,step;
		Iter& operator++(){v+=step;return *this;}
		bool operator!=(Iter& itr){return v<itr.v;}
		int& operator*(){return v;}
	};
	Iter i, n;
	range(int i, int n,int step):i({i,step}), n({n,step}){}
	range(int i, int n):range(i,n,1){}
	range(int n):range(0,n){}
	Iter& begin(){return i;}
	Iter& end(){return n;}
};
struct rrange{
	struct Iter{
		int v,step;
		Iter& operator++(){v-=step;return *this;}
		bool operator!=(Iter& itr){return v>itr.v;}
		int& operator*(){return v;}
	};
	Iter i, n;
	rrange(int i, int n,int step):i({i-1,step}), n({n-1,step}){}
	rrange(int i, int n):rrange(i,n,1){}
	rrange(int n) :rrange(0,n){}
	Iter& begin(){return n;}
	Iter& end(){return i;}
};

//input
template<typename T1,typename T2> istream& operator >> (istream& is,pair<T1,T2>& p){return is>>p.first>>p.second;}
template<typename T1> istream& operator >> (istream& is,tuple<T1>& t){return is >> get<0>(t);}
template<typename T1,typename T2> istream& operator >> (istream& is,tuple<T1,T2>& t){return is >> get<0>(t) >> get<1>(t);}
template<typename T1,typename T2,typename T3> istream& operator >> (istream& is,tuple<T1,T2,T3>& t){return is >>get<0>(t)>>get<1>(t)>>get<2>(t);}
template<typename T1,typename T2,typename T3,typename T4> istream& operator >> (istream& is,tuple<T1,T2,T3,T4>& t){return is >> get<0>(t)>>get<1>(t)>>get<2>(t)>>get<3>(t);}
template<typename T> istream& operator >> (istream& is,vector<T>& as){for(int i:range(as.size()))is >>as[i];return is;}
//output
template<typename T> ostream& operator << (ostream& os, const set<T>& ss){for(auto a:ss){if(a!=ss.begin())os<<" "; os<<a;}return os;}
template<typename T1,typename T2> ostream& operator << (ostream& os, const pair<T1,T2>& p){return os<<p.first<<" "<<p.second;}
template<typename K,typename V> ostream& operator << (ostream& os, const map<K,V>& m){bool isF=true;for(auto& p:m){if(!isF)os<<endl;os<<p;isF=false;}return os;}
template<typename T1> ostream& operator << (ostream& os, const tuple<T1>& t){return os << get<0>(t);}
template<typename T1,typename T2> ostream& operator << (ostream& os, const tuple<T1,T2>& t){return os << get<0>(t)<<" "<<get<1>(t);}
template<typename T1,typename T2,typename T3> ostream& operator << (ostream& os, const tuple<T1,T2,T3>& t){return os << get<0>(t)<<" "<<get<1>(t)<<" "<<get<2>(t);}
template<typename T1,typename T2,typename T3,typename T4> ostream& operator << (ostream& os, const tuple<T1,T2,T3,T4>& t){return os << get<0>(t)<<" "<<get<1>(t)<<" "<<get<2>(t)<<" "<<get<3>(t);}
template<typename T> ostream& operator << (ostream& os, const vector<T>& as){for(int i:range(as.size())){if(i!=0)os<<" "; os<<as[i];}return os;}
template<typename T> ostream& operator << (ostream& os, const vector<vector<T>>& as){for(int i:range(as.size())){if(i!=0)os<<endl; os<<as[i];}return os;}

// values
template<typename T> inline T INF(){assert(false);};
template<> inline int INF<int>(){return 1<<28;};
template<> inline ll INF<ll>(){return 1LL<<58;};
template<> inline double INF<double>(){return 1e16;};
template<> inline long double INF<long double>(){return 1e16;};

template<class T> inline T EPS(){assert(false);};
template<> inline int EPS<int>(){return 1;};
template<> inline ll EPS<ll>(){return 1LL;};
template<> inline double EPS<double>(){return 1e-8;};
template<> inline long double EPS<long double>(){return 1e-8;};

// min{2^r | n < 2^r}
template<typename T> T upper_pow2(T n){ T res=1;while(res<n)res<<=1;return res;}
// max{d | 2^d  <= n}
template<typename T> T msb(T n){ int d=62;while((1LL<<d)>n)d--;return d;}

template<typename T,typename U> T pmod(T v,U M){return (v%M+M)%M;}


template<class Cost> struct Edge{
    int to;Cost cost;
    Edge(int to,Cost cost):to(to),cost(cost){};
    bool operator<(Edge r) const{ return cost<r.cost;}
    bool operator>(Edge r) const{ return cost>r.cost;}
};
template<class Cost> ostream& operator << (ostream& os, const Edge<Cost>& e){ return os<<"(->"<<e.to <<","<<e.cost<<")";}
template<class Cost> using Graph = vector<vector<Edge<Cost>>>;

template<class Cost> vector<Cost> dijkstra(const Graph<Cost>& g,const int s){
    vector<Cost> ds(g.size(),0);ds[s]=INF<ll>();
    priority_queue<pair<Cost,int>,vector<pair<Cost,int>>> que;que.push({INF<ll>(),s});
    while(!que.empty()){
        Cost c;int v;tie(c,v)=que.top();que.pop();
        for(const Edge<Cost>& e:g[v]){
            Cost nxtc= min(c,e.cost);
            if(nxtc > ds[e.to]){
                ds[e.to]=nxtc;
                que.push({nxtc,e.to});
            }
        }
    }
    return ds;
}


ll MOD = 1e9+7;
class Main{
	public:
	void run(){
		int N,M;cin >> N >> M;
		vector<ll> ws(N);cin >> ws;
		vector<pair<int,int>> ps(M); cin >> ps;
		for(int i:range(M))ps[i].first--,ps[i].second--;

		Graph<ll> g(2*N);
		for(int i:range(N)) g[i].push_back({i+N,ws[i]});
		for(int i:range(M)){
			int f,t;tie(f,t)=ps[i];
			g[f+N].push_back({t,INF<ll>()});
		}

		vector<vector<ll>> dp(1050,vector<ll>(1050));
		dp[0][0] = 1;
		for(int i:range(1,1050))for(int j:range(0,1050)){
			// i-1個で既に全射 or i個目で新たに全射
			dp[i][j] = (j*(dp[i-1][j] + dp[i-1][j-1])%MOD)%MOD;
		}
		ll res = 0;
		for(int i:range(N)){
			vector<ll> ds = dijkstra(g,i);
			cerr << ds << endl;
			for(int j:range(N))if( ds[j+N] == ws[j]){
				res = (res + dp[ws[i]][ws[j]])%MOD;
			}
		}
		cout << res << endl;
	}
};

int main(){
	cout <<fixed<<setprecision(20);
	cin.tie(0);
	ios::sync_with_stdio(false);
	Main().run();
	return 0;
}
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