dfs求联通分量

最新推荐文章于 2023-05-17 10:22:22 发布

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最新推荐文章于 2023-05-17 10:22:22 发布

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本文链接：https://blog.youkuaiyun.com/qq_41383801/article/details/87900446

本文探讨了在战争中城市间高速公路被敌军占领后，如何快速计算出需要修复的高速公路数量，以保持剩余城市的连接。通过深度优先搜索算法，计算在移除某城市节点后的连通分量数，进而确定所需的最小修复路线数。

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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 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 Mlines 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

分析：添加的最少的路线，就是他们的连通分量数-1，因为当a个互相分立的连通分量需要变为连通图的时候，只需要添加a-1个路线，就能让他们相连。所以这道题就是求去除了某个结点之后其他的图所拥有的连通分量数
使用邻接矩阵存储，对于每一个被占领的城市，去除这个城市结点，就是把它标记为已经访问过，这样在深度优先遍历的时候，对于所有未访问的结点进行遍历，就能求到所有的连通分量的个数～记得因为有很多个要判断的数据，每一次输入被占领的城市之前要把visit数组置为false

#include<iostream>
#include<string>
#include<stdio.h>
#include<sstream>
#include<math.h>
#include<string.h>
using namespace std;

bool vis[1010];
int a[1010][1010];
int n,m,k;

void dfs(int j)
{
	vis[j]=1;
	for(int i=1;i<=n;i++)
	{
		if(vis[i]==0&&a[i][j]==1) dfs(i);
	}
}

int main()
{
	scanf("%d%d%d",&n,&m,&k);
	memset(a,0,sizeof(a));
	for(int i=0;i<m;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		a[u][v]=1,a[v][u]=1;
	}
	for(int i=0;i<k;i++)
	{
		int t;
		scanf("%d",&t);
		memset(vis,0,sizeof(vis));
		vis[t]=1;
		int cnt=0;
		for(int j=1;j<=n;j++)
		{
			if(vis[j]==0)
			{
				dfs(j);
				cnt++;
			}
		}
		printf("%d\n",cnt-1);
	}
	return 0;
}