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

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思路简述

将城市之间的道路以数组形式保存下来，然后建立城市集合进行等价划分。

利用并查集，每个城市的初始值都赋值为-1，代表各自在一个集合内。

删掉相关道路后，进行union操作，并且每个集合有且仅有一座城市的值为-1

计算-1的个数sum，sum就是集合的个数，集合之间的联通需要sum-1条道路

#include 
using namespace std;
struct road{
    int s,e;
};
int f[1005];
int find(int i){
    if(f[i]<0)return i;
    else return f[i]=find(f[i]);        //压缩路径
}
void union_f(int a,int b){
    if(a==b)return;
    if(f[a]==-1&&f[b]!=-1)swap(a,b);
    f[a]=b;
}
int main(){
    int n,m,k,c;
    cin>>n>>m>>k;
    vectorcity(m);
    for(int i=0;i