PAT.A1013 Battle Over Cities-优快云博客

本文介绍了一种算法，用于确定在某个城市被敌军占领后，为了保持其余城市之间的连接，需要修复多少条高速公路。通过输入城市数量、剩余高速公路数量及关注的城市，算法能够快速计算出每个城市丢失后所需的修复工作。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

#include<iostream>
#include<queue>
#include<map>
#include<string>
#include<cstring>
using namespace std;
const int maxn = 1010;
int n, m, k,sum;
int edge[maxn][maxn] = { 0 }, tmp[maxn][maxn] = { 0 };
bool vist[maxn] = { false };
void dfs(int u) {
	vist[u] = true;
	for (int i = 1; i <= n; i++) {
		if (!vist[i] && tmp[u][i])
			dfs(i);
	}
}
void dfstra() {
	for (int i = 1; i <= n; i++) {
		if (!vist[i]) {
			dfs(i);
			sum++;
		}
	}
}
int main() {
	int a1, a2,b;
	scanf("%d%d%d", &n, &m, &k);
	for (int i = 0; i < m; i++) {
		scanf("%d%d", &a1, &a2);
		edge[a1][a2] = 1;
		edge[a2][a1] = 1;
	}
	for (int i = 0; i < k; i++) {
		scanf("%d", &b);
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= n; j++)
				tmp[i][j] = edge[i][j];
		}
		for (int j = 1; j <=n; j++) {
			vist[b] = true;
			tmp[b][j] = 0;
			tmp[j][b] = 0;
		}
		sum = 0;
		dfstra();
		memset(vist, false, sizeof(vist));
		if (sum > 0) sum -= 1;
		printf("%d\n", sum);
			
	}
	return 0;
}