1013 Battle Over Cities (25 分)

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#c++ #算法

该程序旨在解决战争中城市连接问题。如果某个城市被占领，与之相连的公路将关闭。输入包含城市总数、剩余公路数和需要检查的城市数，以及城市间的公路连接信息。对于每个需要检查的城市，输出失去该城市后需要修复的公路数量以保持其余城市间的连接。例如，在3个城市和2条公路的情况下，如果城市1被占领，则需要修复1条公路（城市2到城市3）。程序通过深度优先搜索计算所需修复的公路数。

在这里插入图片描述
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 city
1

-city
2

and city
1

-city
3

. Then if city
1

is occupied by the enemy, we must have 1 highway repaired, that is the highway city
2

-city
3

.

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:
3 2 3
1 2
1 3
1 2 3
Sample Output:
1
0
0

#include<iostream>
#include<algorithm>
using namespace std;
int edge[1005][1005];
int N;
bool visit[1005];
void dfs(int start) {
	visit[start] = true;
	for (int i = 1; i <= N; i++) {
		if (visit[i] == false && edge[start][i] == 1) {
			dfs(i);
		}
	}
}
int main() {
	int  M, K;
	cin >> N >> M >> K;
	for (int i = 0; i < M; i++) {
		int x, y;
		cin >> x >> y;
		edge[x][y] = edge[y][x] = 1;
	}
	for (int j = 0; j < K; j++) {
		int temp;
		int count = 0;
		fill(visit, visit + 1005, false);
		cin >> temp;
		visit[temp] = true;
		for (int k = 1; k <= N; k++) {
			if (visit[k] == false) {
				dfs(k);
				count++;
			}
		}
		cout << count - 1 << endl;
	}
	return 0;
}