1013 Battle Over Cities （25 分）

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

题意: 给定一个无向图，当删除图中某个结点时，将会把与之相连的边删除，给出k个查询，每个查询给出对应的结点编号，求删除该顶点后需要多少条边才能使图变得连通。

思路:1添加的边数等于连通块个数减1，故题目要求的是连通块的个数。

#include <iostream>
#include <algorithm>
using namespace std;

const int maxn=1010;
const int INF=99999999;
int G[maxn][maxn];
bool vis[maxn];
int deletepoint,n;

void dfs(int v)
{
	if(v==deletepoint){//如果当前结点等于要删除的结点，则return 
		return ;
	} 
	vis[v]=true;
	for(int i=1;i<=n;i++){//从1开始 
		if(vis[i]==false&&G[v][i]==1){
			dfs(i);
		}
	}
}

int main()
{
	int m,k,u,v,num;
	scanf("%d %d %d",&n,&m,&k);
	fill(G[0],G[0]+1010*1010,INF);
	for(int i=0;i<m;i++){
		scanf("%d %d",&u,&v);
		G[u][v]=G[v][u]=1;
	}
	for(int i=0;i<k;i++){
		num=0;
		scanf("%d",&deletepoint);
		fill(vis,vis+1010,false);
		for(int j=1;j<=n;j++){//从1开始 
			if(j!=deletepoint&&vis[j]==false){//要判断是否之前的循环遍历过了 
				dfs(j);
				num++;//放if里面，放外面的话记的是结点的个数 
			}
		}
		printf("%d\n",num-1);
	}
	return 0;
}

标签：25,结点,city,int,Over,highways,maxn,Battle,cities
来源： https://blog.csdn.net/weixin_42105789/article/details/88423194