codeforces 246 D. Colorful Graph (set)

本篇介绍了一个图论问题的解决方法,给定一个无向图,每个顶点都有一个颜色,任务是找到一种颜色,使得该颜色的顶点拥有最多的不同邻接颜色。文章详细解释了输入格式、解决问题的思路及代码实现。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

 D. Colorful Graph

You've got an undirected graph, consisting of n vertices and m edges. We will consider the graph's vertices numbered with integers from 1 to n. Each vertex of the graph has a color. The color of the i-th vertex is an integer ci.

Let's consider all vertices of the graph, that are painted some colork. Let's denote a set of such as V(k). Let's denote the value of theneighbouring color diversity for color k as the cardinality of the set Q(k) = {cu :  cu ≠ k and there is vertex v belonging to set V(k)such that nodes v and u are connected by an edge of the graph}.

Your task is to find such color k, which makes the cardinality of setQ(k) maximum. In other words, you want to find the color that has the most diverse neighbours. Please note, that you want to find such color k, that the graph has at least one vertex with such color.

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105) — the number of vertices end edges of the graph, correspondingly. The second line contains a sequence of integersc1, c2, ..., cn (1 ≤ ci ≤ 105) — the colors of the graph vertices. The numbers on the line are separated by spaces.

Next m lines contain the description of the edges: the i-th line contains two space-separated integers ai, bi (1 ≤ ai, bi ≤ nai ≠ bi) — the numbers of the vertices, connected by the i-th edge.

It is guaranteed that the given graph has no self-loops or multiple edges.

Output

Print the number of the color which has the set of neighbours with the maximum cardinality. It there are multiple optimal colors, print the color with the minimum number. Please note, that you want to find such color, that the graph has at least one vertex with such color.

Example
Input
6 6
1 1 2 3 5 8
1 2
3 2
1 4
4 3
4 5
4 6
Output
3
Input
5 6
4 2 5 2 4
1 2
2 3
3 1
5 3
5 4
3 4
Output
2

n个顶点,各有一个颜色,可以相同,m条边,问相邻的颜色最多的颜色号码是多少

code:

#include <iostream>
#include<cstdio>
#include<set>
#include<cstring>
using namespace std;
set<int>s[100010];
int co[100010];
bool vis[100010];
int main()
{
    int n,m,x,y;
    scanf("%d%d",&n,&m);
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n;i++)
    {
        s[i].clear();
        scanf("%d",&co[i]);
        vis[co[i]]=1;
    }
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d",&x,&y);
        {
            if(co[x]!=co[y])
            {
                s[co[x]].insert(co[y]);
                s[co[y]].insert(co[x]);
            }
        }
    }
    int maxn=-1,ans;
    for(int i=1;i<100001;i++)
    {
        if(vis[i]&&maxn<(int)s[i].size())
        {
            maxn=(int)s[i].size();
            ans=i;
        }
    }
    printf("%d\n",ans);
    return 0;
}


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值