codeforces244C. Checkposts

强连通分量模板题

/* * Tarjan 算法 * 复杂度O(m+n); */
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define inf -0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define mem(a, b) memset(a, b, sizeof(a))
typedef long long ll;
const int maxn=500001;
const int MOD=1e9+7;
vector<int>G[maxn];
int Low[maxn],Stack[maxn],DFN[maxn],Belong[maxn];//Belong数组的值是1-scc;
int Index,top;
int a[maxn];
int scc;//强连通分量的个数
int cost[maxn];
bool Instack[maxn];
int num[maxn]; //各个强连通分量包含的点的个数,数组编号1-scc
//num数组不一定需要,结合实际情况

void Tarjan(int u){
    int v;
    Low[u]=DFN[u]=++Index;
    Stack[top++]=u;
    Instack[u]=true;
    for(int i=0;i<G[u].size();i++){
        v=G[u][i];
        if(!DFN[v]){
            Tarjan(v);
            Low[u]=min(Low[u],Low[v]);
        }
        else if(Instack[v]&&Low[u]>DFN[v])
            Low[u]=DFN[v];
    }
    if(Low[u]==DFN[u]){
        scc++;
        cost[scc]=INF;
        do{
            v=Stack[--top];
            Instack[v]=false;
            Belong[v]=scc;
            if(cost[scc]==a[v]){
                num[scc]++;
            }
            else if(cost[scc]>a[v]){
                num[scc]=1;
                cost[scc]=a[v];
            }
        }while(v!=u);
    }
    return ;
}

void solve(int n){
    mem0(DFN);
    memset(Instack,false,sizeof(Instack));
    mem0(num);
    Index=scc=top=0;
    for(int i=1;i<=n;i++)
        if(!DFN[i])
            Tarjan(i);
}

void init(int n){
    for(int i=1;i<=n;i++)
        G[i].clear();
}

int main(){
    int n,m;
    while(scanf("%d",&n)!=EOF){
        init(n);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        scanf("%d",&m);
        int x,y;
        for(int i=0;i<m;i++){
            scanf("%d%d",&x,&y);
            G[x].push_back(y);
        }
        solve(n);
        ll ans1=0,ans2=1;
        for(int i=1;i<=scc;i++){
            ans1+=cost[i];
            ans2=ans2*num[i]%MOD;
        }
        printf("%I64d %I64d\n",ans1,ans2);
    }
    return 0;
}