PAT 1013 Battle Over Cities

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

思路

这题考察的就是并查集，对于每个要删除的城市：首先初始化并查集，然后建立并查集，遇到要删除的那个点就跳过，最后统计并查集中集合的个数cnt，最后cnt-2就是答案。因为cnt个集合，所以全部合并需要cnt-1条连线，又因为删除了1个点，导致多了一个没用的集合，所以应该把剩下的cnt-1个集合全部合并，所以最终结果为cnt-2。

注意这题使用cin会超时，建议以后都用scanf，scanf的注意点：https://www.cnblogs.com/FengZeng666/p/12609218.html

#include <iostream>
#include <stdio.h>


using namespace std;

struct Edge
{
    int a;
    int b;
}edge[1100000];

int Tree[1100];

int findRoot(int x)
{
    if(Tree[x] == -1)
        return x;
    int tmp = findRoot(Tree[x]);
    Tree[x] = tmp;
    return tmp;
}


int main()
{
    int n, m, k;
    scanf("%d%d%d", &n, &m, &k);
    
    for(int i = 0; i < m; ++i)
        scanf("%d%d", &edge[i].a, &edge[i].b);
    
    int t;
    for(int x = 1; x <= k; ++x)
    {
        scanf("%d", &t);
        
        // 初始化并查集 
        for(int i = 1; i <= n; ++i)
            Tree[i] = -1;
        
        // 遍历所有边，建立并查集 
        for(int i = 0; i < m; ++i)
        {
            if(edge[i].a == t || edge[i].b == t)
                continue;
                
            int ra = findRoot(edge[i].a);
            int rb = findRoot(edge[i].b);
            // 合并 
            if(ra != rb)
                Tree[ra] = rb;
        }
        
        int cnt = 0;    //并查集中集合的个数 
        for(int i = 1; i <=n ;++i)
        {
            if(Tree[i] == -1)
                cnt++; 
        }
        
        if(x == k)
            cout << cnt-2;
        else
            cout << cnt-2 << endl;
    }
    
    
    return 0;
}