PAT甲级1013 Battle Over Cities（简单图论）

最新推荐文章于 2025-08-10 22:40:20 发布

threeh20

最新推荐文章于 2025-08-10 22:40:20 发布

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本文介绍了一个算法挑战，即在战争中保持城市间高速公路连接的重要性。当某个城市被敌方占领，所有通往该城市的道路将关闭，我们需要立即知道是否需要修复其他道路以保持其余城市间的连接。通过提供地图和城市间的剩余道路，文章展示了如何计算在特定城市失去后，需要修复的道路数量。

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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 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 Knumbers, 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

类似于数连通块，某一点断了，就把与该点相关联的边去掉。然后数连通块。答案是连通块数目-1

#include<bits/stdc++.h>
using namespace std;
bool vis[1111];
vector<int> p[1111];
void del(int x)
{
	for(int i=0;i<p[x].size();i++)
	{
		if(vis[p[x][i]]==0)
		{
			vis[p[x][i]]=1;
			del(p[x][i]);
		}
	}
}
int main(){
	int n,m,k;
	cin>>n>>m>>k;
	for(int i=1;i<=n;i++)
	p[i].clear();
	int a,b;
	for(int i=0;i<m;i++)
	{
		scanf("%d%d",&a,&b);
		p[a].push_back(b);
		p[b].push_back(a);
	}
	int skr;
	while(k--)
	{
		scanf("%d",&skr);
		memset(vis,0,sizeof(vis));
		vis[skr]=1;
		int js=0;
		for(int i=1;i<=n;i++)
		{
			if(vis[i]==0)
			{
				js++;
				vis[i]=1;
				del(i);
			}
		}
		cout<<js-1<<endl;
	}
	return 0;
}