【Codeforces 427C】Checkposts

环里面的点只需要一个点就能全都保护
问你最少需要多少花费以及最少的点才能将所有的点都保护

【题解】

有向图的强连通分量求出所有的联通分量
显然每个联通分量里面只需选择最小那个点就好
如果有多个最小的点,那么这个环就有多个选择。
每个环的最小点个数连乘一下就是方案数了
然后每个联通分量的最小值再全部都加起来就ok

【代码】

import java.io.*;

import java.util.*;

public class Main {

    static InputReader in;

    static PrintWriter out;

    public static void main(String[] args) throws IOException{

        //InputStream ins = new FileInputStream("E:\\rush.txt");

        InputStream ins = System.in;

        in = new InputReader(ins);

        out = new PrintWriter(System.out);

        //code start from here

        new Task().solve(in, out);

        out.close();

    }

    static int N = (int)1e5;

    static class Task{

        int n,m;

        ArrayList g[] = new ArrayList[N+10],g1[] = new ArrayList[N+10];

        int cost[] = new int[N+10];

        int dfn[] = new int[N+10],low[] = new int[N+10];

        boolean in[] = new boolean[N+10];

        int mystack[] = new int[N+10];

        int tot = 0,top = 0,totn = 0;

        void dfs(int x) {

        	dfn[x] = low[x] = ++tot;

        	mystack[++top] = x;

        	in[x] = true;

        	int len = g[x].size();

        	for (int i = 0;i < len;i++) {

        		int y = (int)g[x].get(i);

        		if (dfn[y]==0) {

        			dfs(y);

        			low[x] = Math.min(low[x],low[y]);

        		}else

        			if (in[y] && dfn[y]<low[x]){

        				low[x] = dfn[y];

        			}

        	}

        	if (low[x]==dfn[x]) {

        		int v = 0;

        		totn++;

        		while (v!=x) {

        			v = mystack[top];

        			in[v] = false;

        			g1[totn].add(v);

        			top--;

        		}

        	}

        }

        public void solve(InputReader in,PrintWriter out) {

        	for (int i = 1;i <= N;i++) {

        		g[i] = new ArrayList();

        		g1[i] = new ArrayList();

        	}

        	n = in.nextInt();

        	for (int i = 1;i <= n;i++) cost[i] = in.nextInt();

        	m = in.nextInt();

        	for (int i = 1;i <= m;i++) {

        		int x,y;

        		x = in.nextInt();y = in.nextInt();

        		g[x].add(y);

        	}

        	for (int i = 1;i <= n;i++)

        		if (dfn[i]==0) dfs(i);

        	long ans = 0;

        	long way = 1;

        	for (int i = 1;i <= totn;i++) {

        		long mi = cost[(int)g1[i].get(0)];

        		long cnt = 1;

        		for (int j = 1;j < (int)g1[i].size();j++) {

        			int x = (int)g1[i].get(j);

        			x = cost[x];

        			if (x<mi) {

        				mi = x;

        				cnt = 1;

        			}else if (x==mi) cnt++;

        		}

        		ans = ans + mi;

        		way = (way*cnt)%((int)1e9 + 7);

        	}

        	out.println(ans+" "+way);

        }

    }

    static class InputReader{

        public BufferedReader br;

        public StringTokenizer tokenizer;

        public InputReader(InputStream ins) {

            br = new BufferedReader(new InputStreamReader(ins));

            tokenizer = null;

        }

        public String next(){

            while (tokenizer==null || !tokenizer.hasMoreTokens()) {

                try {

                tokenizer = new StringTokenizer(br.readLine());

                }catch(IOException e) {

                    throw new RuntimeException(e);

                }

            }

            return tokenizer.nextToken();

        }

        public int nextInt() {

            return Integer.parseInt(next());

        }

    }

}

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