1013 Battle Over Cities (25 point(s))

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在战争背景下，确保城市间高速公路连通至关重要。本文介绍了一种算法，用于快速计算若某城市被敌方占领，需要修复多少条高速公路以保持剩余城市间的连接。通过并查集和连通分支数的概念，文章提供了一个示例输入输出，并详细解释了处理步骤。

1013 Battle Over Cities (25 point(s))

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:

Sample Output:

1
0
0

思考：

题意：城市和高速公路构成带权无向图，现在去掉一些城市，其删点子图需要多少条道路使得相互连通。

知识点：并查集，连通分支数

注意点：

（1）findRoot(x)函数；

（2）由于有多个提问，因此需要存储所有高速公路端点（玄学TLE）；

（3）注意需要对于删点子图的处理，即遇到该城市（点）或者包含该城市的高速公路（边）都要跳过；

（4）最终需要新修建连通分支数-1条道路。

#include<iostream>
#include<vector>
#include<string>
using namespace std;
const int MAX = 1e3+7;
int Tree[MAX];
int findRoot(int x){
	if(Tree[x]==-1) return x;
	else{
		int temp = findRoot(Tree[x]);
		Tree[x] = temp;
		return temp;
	}
}
struct highway{
	int a,b;
}; 
int main(void){
	int N,M,K;//cities,highways,to be checked
	cin>>N>>M>>K;
	vector<highway> p;
	highway hw;
	for(int i=0;i<M;i++){
		cin>>hw.a>>hw.b;
		p.push_back(hw);
	}
	int t;// what we concern?
	for(int k=0;k<K;k++){
		cin>>t;
		for(int i=1;i<=N;i++) Tree[i]=-1;
		for(int i=0;i<M;i++){
			int c1 = p[i].a;
			int c2 = p[i].b;
			if(c1==t||c2==t) continue;
			c1 = findRoot(c1);
			c2 = findRoot(c2);
			if(c1!=c2){
				Tree[c1]=c2;
			} 
		}
		int branch=0;
		for(int i=1;i<=N;i++){
			if(i==t) continue;
			if(Tree[i]==-1) branch++;
		}
		cout<<branch-1<<endl;
	} 
	return 0;
}