【Codeforces 489D】Unbearable Controversy of Being

【链接】 我是链接,点我呀:)

【题意】

让你找到(a,b,c,d)的个数
这4个点之间有4条边有向边
(a,b)(b,c)
(a,d)(d,c)
即有两条从a到b的路径,且这两条路径分别经过b和d到达c

【题解】

我们枚举a,c
然后找到这样的b的个数cntb,其中a到b有一条边,b到c也有一条边
显然从这cntb个b中取2个
就能组成b和d了。
即C(N,2)
怎么得到这样的b的个数呢
先枚举a的所有出点v,然后看看v是否和c联通即可.
联通的话,cntb++

【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 = 3000;
    static class Task{

        int n,m;
        ArrayList<Integer> g[] = new ArrayList[N+10];
        boolean can[][] = new boolean[N+10][N+10];

        public void solve(InputReader in,PrintWriter out) {
        	for (int i = 1;i <= N;i++) {
        		g[i] = new ArrayList<Integer>();
        	}
        	n = in.nextInt(); m = in.nextInt();
        	for (int i = 1;i <= m;i++) {
        		int x,y;
        		x = in.nextInt();y = in.nextInt();
        		can[x][y] = true;
        		g[x].add(y);
        	}
        	long ans = 0;
        	for (int i = 1;i <= n;i++)
        		for (int j = 1;j <= n;j++)
        			if (i!=j) {
        				long numberofpath2 = 0;
        				for (int l = 0;l < (int)g[i].size();l++) {
        					int y = g[i].get(l);
        					if (can[y][j]) {
        						numberofpath2++;
        					}
        				}
        				ans = ans + numberofpath2*(numberofpath2-1)/2;
        			}
        	out.println(ans);
        }
    }

    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());
        }
    }
}