P2194 HXY烧情侣(tarjan&组合数学)

Posted 2022-04-30 Harris-H

P2194 HXY烧情侣(tarjan&组合数学)

显然就是对每个强连通分量 找到费用最小的点的个数。

$\prod cn{t}_{w_{mi{n}_i}}$ 就是答案。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull; 
const int N=1e5+5,M=3e5+5,inf=0x3f3f3f3f,mod=1e9+7;
const int hashmod[4] = 402653189,805306457,1610612741,998244353;
#define mst(a,b) memset(a,b,sizeof a)
#define db double
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define x first
#define y second
#define pb emplace_back
#define SZ(a) (int)a.size()
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define ios ios::sync_with_stdio(false),cin.tie(nullptr) 
void Print(int *a,int n)
	for(int i=1;i<n;i++)
		printf("%d ",a[i]);
	printf("%d\\n",a[n]); 

template <typename T>		//x=max(x,y)  x=min(x,y)
void cmx(T &x,T y)
	if(x<y) x=y;

template <typename T>
void cmn(T &x,T y)
	if(x>y) x=y;

//tarjan O(n+m)
int n,m,h[N],cnt;
int dfn[N],low[N],id,vis[N];
int w[N];
//bool cut[N]; judge cut point
stack<int>s;
struct edge
	int to,nt;
e[M];
void add(int u,int v)
	e[++cnt]=v,h[u],h[u]=cnt;

ll ans,res=1;
void dfs(int u)
	low[u]=dfn[u]=++id;vis[u]=1;
	s.push(u);
	//int son=0;
	for(int i=h[u];i;i=e[i].nt)
		int v=e[i].to;
		if(!dfn[v])
			dfs(v);
			low[u]=min(low[u],low[v]);
			//if(low[v]>=dfn[u]) cut[u]=1;
		else if(vis[v]) low[u]=min(low[u],dfn[v]);
	
	if(dfn[u]==low[u])	//shrink point

		vis[u]=0;
		int mn = w[u];
		int js = 1;
		while(s.top()!=u)
			if(mn > w[s.top()])
				mn = w[s.top()];
				js = 1;
			
			else if(mn == w[s.top()])
				js++;
			
			vis[s.top()]=0;
			s.pop();
		
		ans+=mn;
		res=res*js%mod;
		s.pop();
	
	//if(u==rt&&son==1) cut[u]=0;

int main()
	scanf("%d",&n);
	rep(i,1,n) scanf("%d",&w[i]);
	scanf("%d",&m);
	rep(i,1,m)
		int u,v;
		scanf("%d%d",&u,&v);
		add(u,v);
	
	rep(i,1,n) if(!dfn[i]) dfs(i);
	printf("%lld %lld\\n",ans,res);
	return 0;