1013 Battle Over Cities 甲级 xp_xht123

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 city1​-city2​ and city1​-city3​. Then if city1​ is occupied by the enemy, we must have 1 highway repaired, that is the highway city2​-city3​.

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

 解题思路：标准的并查集，保持其余城市连通性，需要维修的最少高速公路条数

标准find函数

int find(int x)
{
    if(p[x] != x) p[x] = find(p[x]);
    return p[x];
}

注意每次判断的时候需要将p数组初始化

判断的方法就是每一个输入的数，与原先输入的联通的边的节点对比，如果发现这个数没有出现在某一条边的两个节点中，那么证明这个节点并不与这两个节点相连，需要再判断一下祖宗节点是否一样，不同证明拥有两个联通块。

#include
#include
 
using namespace std;
const int N = 10100 , M = 500010;
int p[N] , n , m , k;
struct edge
{
    int a , b;
}e[M];
 
int find(int x)
{
    if(p[x] != x) p[x] = find(p[x]);
    return p[x];
}
 
int main()
{
    scanf("%d %d %d", &n ,&m ,&k);
    
    for(int i = 0;i < m;i ++) scanf("%d %d", &e[i].a, &e[i].b);
    
    while(k --)
    {
        int num;
        scanf("%d", &num);
        
        for(int i = 1;i <= n;i ++) p[i] = i;
        
        int cnt = n - 1;//最多连n-1条边
        
        for(int i = 0;i < m;i ++)
        {
            int ea = e[i].a , eb = e[i].b;
            if(ea != num && eb != num)//num这个城市和ea、eb城市不相连
            {
                int pa = find(ea) , pb = find(eb);
                
                if(pa != pb) p[pa] = pb , cnt --;//连接两个连通块
            }
        }
        
        printf("%d\n" , cnt - 1);//注意最后需要减一排除重连的情况
    }
    return 0;
}