1013. Battle Over Cities (25)（C++)

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最新推荐文章于 2024-08-07 14:44:24 发布

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分类专栏： PAT甲级

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本文介绍了一种算法，用于计算当某个城市被敌军占领后，为了保持剩余城市的连接，需要修复多少条高速公路。通过使用深度优先搜索(DFS)或并查集方法，可以快速得出答案。

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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

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

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

题目大意：去掉一个连通图中的一个节点，剩下的节点还能构成连通图吗，不能的话，最少需要几条路将其联系在一起？（其实就是求连通图被分成了几块，即连通分量的个数）。

解题思路：直接遍历图就好了，就直接当那个丢失的点已经被访问过了，dfs最方便了，嘻嘻——

DFS

#include <iostream>
#include <vector>
using namespace std;
int n, m, k;
vector<int>road[1005];
bool visit[1005];
void dfs(int index){
	visit[index] = true;
	for(int i = 0; i < road[index].size(); ++ i)
		if(visit[road[index][i]] == false)
			dfs(road[index][i]);
}
int main(){
	scanf("%d %d %d",&n,&m,&k);
	for(int i = 0; i < m; ++ i){
		int a, b;
		scanf("%d %d",&a,&b);
		road[a].push_back(b);
		road[b].push_back(a);
	}
	for(int i = 0; i < k; ++ i){
		fill(visit, visit + 1005, false);
		int temp, cnt = -1;
		scanf("%d",&temp);
		visit[temp] = true;
		for(int i = 1; i <= n; ++ i){
			if(visit[i] == false){
				++ cnt;
				dfs(i);
			}
		}
		printf("%d\n", cnt);
	}
}

并查集

#include <iostream>
#include <vector>
using namespace std;
struct Edge
{
    int from, to;
};
int n, m, k, temp;
int father[1005];
int findfather(int x){
    if(x == father[x])
        return x;
    int temp = findfather(father[x]);
    father[x] = temp;
    return temp;
}
int countFather(){
    int cnt = -2;
    for(int i = 1; i <= n; ++ i)
        if(i == father[i])
            cnt += 1;
    return cnt;
}
int main(){
    scanf("%d %d %d",&n,&m,&k);
    vector<Edge>edges(m);
    for(int i = 0; i < m; ++ i)
        scanf("%d %d",&edges[i].from,&edges[i].to);
    for(int i = 0; i < k; ++ i){
        for(int j = 1; j <= n; ++ j)
            father[j] = j;
        scanf("%d", &temp);
        for(Edge e : edges){
            int ua = findfather(e.from), ub = findfather(e.to);
            if(ua != temp && ub != temp && ua != ub)
                father[ua] = ub;
        }
        printf("%d\n", countFather());
    }
}