【链接】 我是链接,点我呀:)
【题意】
【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());
}
}
}