1013 Battle Over Cities-PAT甲级-优快云博客

本文探讨了在战争背景下，确保城市间高速公路网络连通性的算法。通过删除被敌方占领城市的连接，算法快速计算出为保持剩余城市连通所需修复的高速公路数量。示例展示了如何在给定城市和高速公路地图的情况下，高效地解决这一问题。

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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

题意分析
给定一个无向图，当删除其中一个结点时，会把跟这个结点有关的边全部删除，问要保持其余的顶点依旧连通，至少需要增加几条边

算法设计
当删除其中一个顶点及其相关的边之后，计算出剩下的图的连通分量，那么增加的边就应该是求出的连通分量-1。

满分代码如下：

#include<bits/stdc++.h>
using namespace std;
const int N=1005;
vector<int>graph[N];
int fa[N];
void init(){
	for(int i=0;i<N;i++)
		fa[i]=i;
}
int get(int x){
	if(fa[x]==x)
		return fa[x];
	return fa[x]=get(fa[x]);
}
void merge(int x,int y){
	x=get(x);
	y=get(y);
	if(x!=y)
		fa[x]=y;
}
int n,m,k;
int main(){
	scanf("%d%d%d",&n,&m,&k);
	int u,v;
	for(int i=1;i<=m;i++){
		scanf("%d%d",&u,&v);
		graph[u].push_back(v);
		graph[v].push_back(u);
	}
	for(int i=1;i<=k;i++){
		init();
		int x,num=0;
		scanf("%d",&x);
		for(int j=1;j<=n;j++){
			if(j==x) continue;
			for(auto h:graph[j]){
				if(h==x) continue;
				merge(j,h);
			}
		}
		for(int j=1;j<=n;j++){
			if(j==x) continue;
			if(fa[j]==j) num++;
		}
		printf("%d\n",num-1);
	}
	return 0;
}