【Codeforces 246D】Colorful Graph

【链接】 我是链接,点我呀:)
【题意】

让你找到所有和x颜色的点中,和该颜色的点颜色不同的相邻的点的个数(重复颜色算一次) 求出哪种颜色的所要求的点的数量最多.

【题解】

对于每一条边只会被查到两次。 所以按照题意暴力枚举的时间复杂度就是O(n+m)级别的 至于查重 只要对找过的点打个标记就很好处理的

【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;
    	int c[] = new int[N+10];
    	ArrayList<Integer> g[] = new ArrayList[N+10];
    	ArrayList<Integer> col[] = new ArrayList[N+10];
    	int mark[] = new int[N+10];
    	
        public void solve(InputReader in,PrintWriter out) {
        	for (int i = 1;i <= N;i++) g[i] = new ArrayList<Integer>();
        	for (int i = 1;i <= N;i++) col[i] = new ArrayList<Integer>();
        	n = in.nextInt();m = in.nextInt();
        	for (int i = 1;i <= n;i++) {
        		c[i] = in.nextInt();
        		col[c[i]].add(i);
        	}
        	for (int i = 1;i <= m;i++) {
        		int x,y;
        		x = in.nextInt();y = in.nextInt();
        		g[x].add(y);g[y].add(x);
        	}
        	int ans = -1,idx = 0;
        	for (int i = 1;i <= N;i++)
        		if (!col[i].isEmpty()) {
        			int cnt = 0;
        			for (int J = 0;J < (int)col[i].size();J++) {
        				int x = col[i].get(J);
        				int len = g[x].size();
                		for (int j = 0;j < len;j++) {
                			int y = g[x].get(j);
                			if (c[y]!=c[x]) {
                				if (mark[c[y]]!=i) {
                					cnt++;
                					mark[c[y]] = i;
                				}
                			}
                		}
        			}
            		if (cnt>ans) {
            			ans = cnt;
            			idx = i;
            		}
        		}
        	out.println(idx);
        }
    }

    

    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());
        }
    }
}
posted @ 2019-03-08 11:01  AWCXV  阅读(253)  评论(0编辑  收藏  举报