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

Solution Code:

• 使用cin輸入存在超時問題，因爲對於大量輸入操作，cin有同步機制，所以推薦使用scanf(或者禁用cin同步)。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

/*************************
用鄰接矩陣存儲城市和道路之間的關係。

該問題可轉化爲：
刪除連通圖中的一個節點後，需要多少條邊，使得其他節點再次成爲連通圖。
需要最少邊數 = 連通分量 - 1。

由於修改矩陣過於繁瑣，可以設被攻佔的城市被訪問過。
**************************/

// 矩陣表示城市間道路
int roads[1001][1001] = {0}; //無窮大
int visited_citys[1001] = {0}; //未被訪問過

void dfs(int size, int begin_city)
{
	visited_citys[begin_city] = 1;
	for (int i = 1; i <= size; ++i) {
		if (roads[begin_city][i] == 1 && visited_citys[i] == 0) {
			dfs(size, i);
			break;
		}
	}
}

int getRepaired(int size, int city)
{
	int connected_component = 0; //連通分量

	memset(visited_citys, 0, sizeof(visited_citys));
	visited_citys[city] = 1;

	// 尋找連通分量
	for (int i = 1; i <= size; ++i) {
		if (visited_citys[i] == 0){
			dfs(size, i);
			connected_component++;
		} 
	}

	return connected_component - 1;
}

int main()
{
	// n - the total number of cities
	// m - the number of remaining highways
	// k - the number of cities to be checked
	// ios::sync_with_stdio(false); //禁用cin同步
	int n, m, k;
	scanf("%d %d %d", &n, &m, &k);

	for (int i = 0; i < m; ++i){
		int begin, end;
		scanf("%d %d", & begin, &end);
		roads[begin][end] = 1;
		roads[end][begin] = 1;
	}

	int city, repaired_roads;
	for (int i = 0; i < k; ++i){
		scanf("%d", &city);
		repaired_roads = getRepaired(n, city);
		cout << repaired_roads << endl;
	}

	return 0;
}