1013 Battle Over Cities （25 分）

It is vitally important to have all the cities connected by highways in a war. If a city is occupied by the enemy, all the highways from/toward that city are closed. We must know immediately if we need to repair any other highways to keep the rest of the cities connected. Given the map of cities which have all the remaining highways marked, you are supposed to tell the number of highways need to be repaired, quickly.

For example, if we have 3 cities and 2 highways connecting city​1​​-city​2​​ and city​1​​-city​3​​. Then if city​1​​ is occupied by the enemy, we must have 1 highway repaired, that is the highway city​2​​-city​3​​.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 3 numbers N (<1000), M and K, which are the total number of cities, the number of remaining highways, and the number of cities to be checked, respectively. Then M lines follow, each describes a highway by 2 integers, which are the numbers of the cities the highway connects. The cities are numbered from 1 to N. Finally there is a line containing K numbers, which represent the cities we concern.

Output Specification:

For each of the K cities, output in a line the number of highways need to be repaired if that city is lost.

Sample Input:

3 2 3
1 2
1 3
1 2 3

Sample Output:

1
0
0

題中要求為如果刪除某個節點，與這個節點連線的所有道路全部不通，要求計算出最少需要增加多少條路使圖變連通

連通的話，即只含有一個連通分量，減少一個節點，那麼不次此節點算在內，之後計算出連通分量的個數num，num-1即為最少新增道路的條數

因為為無向連通圖，相對於有向連通圖簡單了很多

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define ll long long
using namespace std;
const int maxn = 1100;
int n,m,k;
bool mapp[maxn][maxn],used[maxn];

void dfs(int pos)
{
    for(int i = 1;i <= n;i ++)
        if(!used[i] && mapp[pos][i])
        {
            used[i] = 1;
            dfs(i);
        }
}
int main()
{
    int x,y,z,_num = 0;
    scanf("%d%d%d",&n,&m,&k);
    for(int i = 0;i < m;i ++)
    {
        scanf("%d%d",&x,&y);
        mapp[x][y] = 1,mapp[y][x] = 1;
    }
    for(int i = 0;i < k;i ++)
    {
        _num = 0;
        scanf("%d",&z);
        memset(used,0,sizeof(used));
        used[z] = true;
        for(int j = 1;j <= n;j ++)
            if(!used[j])
                used[j] = 1,dfs(j),_num++;
        printf("%d\n",max(0,_num));
    }
    return 0;
}