Hongcow Builds A Nation ___codeforces 744A

题目

https://cn.vjudge.net/problem/592815/origin

Hongcow is ruler of the world. As ruler of the world, he wants to make it easier for people to travel by road within their own countries.

The world can be modeled as an undirected graph with n nodes and m edges. k of the nodes are home to the governments of the k countries that make up the world.

There is at most one edge connecting any two nodes and no edge connects a node to itself. Furthermore, for any two nodes corresponding to governments, there is no path between those two nodes. Any graph that satisfies all of these conditions is stable.

Hongcow wants to add as many edges as possible to the graph while keeping it stable. Determine the maximum number of edges Hongcow can add.

Input

The first line of input will contain three integers n, m and k (1 ≤ n ≤ 1 000, 0 ≤ m ≤ 100 000, 1 ≤ k ≤ n) — the number of vertices and edges in the graph, and the number of vertices that are homes of the government.

The next line of input will contain k integers c₁, c₂, ..., c_k (1 ≤ c_i ≤ n). These integers will be pairwise distinct and denote the nodes that are home to the governments in this world.

The following m lines of input will contain two integers u_i and v_i (1 ≤ u_i, v_i ≤ n). This denotes an undirected edge between nodes u_i and v_i.

It is guaranteed that the graph described by the input is stable.

Output

Output a single integer, the maximum number of edges Hongcow can add to the graph while keeping it stable.

Example

Input

4 1 2
1 3
1 2

Output

2

Input

3 3 1
2
1 2
1 3
2 3

Output

0

Note

For the first sample test, the graph looks like this:

Vertices  1 and  3 are special. The optimal solution is to connect vertex  4 to vertices  1 and  2. This adds a total of  2 edges. We cannot add any more edges, since vertices  1 and  3 cannot have any path between them.

For the second sample test, the graph looks like this:

We cannot add any more edges to this graph. Note that we are not allowed to add self-loops, and the graph must be simple.

思路

    这个题算是一个贪心算法的题目，但是贪心的算法本身很多变，很能简单的说明白具体应该怎么做，就和dp一样有时很难定义dp的量，

代码

#include <stdio.h>
#include <vector>
#include <stack>
#include <algorithm>
#include <map>
using namespace std;
vector<int> g[1001];
int node[1001];
int main(){
    int n, m, k, temp, res = -1;
    scanf("%d%d%d", &n, &m, &k);
    stack<int> s, stem;
    for(int i = 1; i <= k; i++){
        scanf("%d", &temp);
        s.push(temp);
    }
    int a,b;
    for(int i = 1; i <= m; i++){
        scanf("%d%d", &a, &b);
        g[a].push_back(b);
        g[b].push_back(a);
    }
    int poi = 0, index = 0, ki = 1;
    vector<int> vectem;
    if(s.empty())
        res = n * (n-1) / 2 - m;
    while(!s.empty()){
        poi = 0;vectem.clear();
        temp = s.top();s.pop();
        stem.push(temp);
        vectem.push_back(temp);
        poi++;
        while(!stem.empty()){
            temp = stem.top();
            stem.pop();
            for(int i = 0; i < g[temp].size(); i++){
                if(find(vectem.begin(), vectem.end(), g[temp][i]) == vectem.end()){
                    vectem.push_back(g[temp][i]);
                    stem.push(g[temp][i]);
                    poi++;
                }
            }
        }
        node[ki++] = (poi - 1) * poi / 2;
        n -= poi;
        if(poi > res) {res = poi;index = ki-1;}
    }
    int sum = 0;
    for(int i = 1; i <= k; i++){
        if(i != index)
            sum += node[i];
        else 
            sum += (res + n) * (res + n - 1) / 2;
    }
    printf("%d", sum - m);
    return 0;
}