PAT甲级1013

PAT甲级1013

原题：
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-city​2 and city1-city​3. Then if city1 is occupied by the enemy, we must have 1 highway repaired, that is the highway city2-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

题目大意：给出n个城市，m条公路，待查城市k座。接下来给出m行数据，代表连接两个城市的公路，现假设某个城市被丧尸攻占，那所有与之相连的公路都不可用，要求需要修复几条路才能使剩下的城市全部连通。
基本上就是求连通图的问题，而且并不要求具体连接哪几座，只需把被攻陷的城市去掉后剩下的连通数减1即可，题目并没有说城市数量和公路一定大于1，也有可能只有1座城市一条公路都没有的情况，这时候不能输出-1

/*
4 4 4
1 2
1 4
2 3
3 4
1 2 3 4
 */
#include <iostream>
#include <vector>
#include <algorithm>
#define MAXN 1001
using namespace std;

int n,m,k,cnt=0;
int highway[MAXN][MAXN];
bool visit[MAXN]={false};
void DFS(int index){
	if(visit[index]==true) return ;
	visit[index]=true;
	for(int i=1;i<=n;i++){
		if(highway[index][i]>0&&!visit[i]) DFS(i);
	}
}
void DFSTraversal(int lost){
	visit[lost]=true;	//该城市已经沦陷，与之相连的所有公路全部断掉
	for(int i=1;i<=n;i++){
		if(i!=lost&&!visit[i]){	//当前访问的不是沦陷城市并且还未访问过
			visit[i]=true;
			for(int j=1;j<=n;j++){
				if(highway[i][j]>0&&!visit[j]&&j!=lost) DFS(j);	//若从i到j有公路且j未访问过并且j未沦陷
			}
			cnt++;
		}
	}
}
int main(){
	int city1,city2;
	scanf("%d%d%d",&n,&m,&k);
	int query[k];
	for(int i=0;i<m;i++){
		scanf("%d%d",&city1,&city2);
		highway[city1][city2]=1;
		highway[city2][city1]=1;
	}
	for(int i=0;i<k;i++) scanf("%d",&query[i]);
	for(int i=0;i<k;i++){
		cnt=0;
		fill(visit,visit+MAXN,false);
		DFSTraversal(query[i]);
		printf("%d\n",cnt==0?0:cnt-1);
	}
	system("pause");
	return 0;
}