【PTA】1013 Battle Over Cities （圖的dfs+統計連通分支數目）

本題考點：

• 統計圖的強連通分支數目

題目：

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 city1-city3.Then if city​1 is occupied by the enemy, we must have 1 highway repaired, that is the highway city​2-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:
3 2 3
1 2
1 3
1 2 3
Sample Output:
1
0
0

本題需要的大意是去除掉圖中某點之後，計算出剩餘點的連通分支數目。

所以採用鄰接矩陣（小於1000個點基本都沒問題）保存圖，然後使用 DFS 來遍歷圖，計算出所有的連通分支數目。

關於圖的 dfs 和 bfs ，之前也做過類似的題目：PTA 7-33 地下迷宮探索 （圖的DFS）。

下面給出實現代碼：

/* 
    統計連通分支總數
    鄰接矩陣 + dfs
 */
#include <iostream>
#include <vector>
#define LOCAL
using namespace std;

const int maxn = 1010;

int N, M, K;    // 城市總數，道路總數，需要檢查的城市總數
int G[maxn][maxn];
bool vis[maxn]; // 看是否被訪問過
int checkedCity;

void dfs(int city)
{
    for (int i = 1; i <= N; i++)
    {
        if(G[city][i] == 1 && vis[i] == false && i != checkedCity)
        {
            vis[i] = true;
            dfs(i);
        }
    }
}


int dfsTravel(int city)
{
    fill(vis, vis + maxn, false);   // 初始化爲未訪問
    int ans = 0;
    for (int i = 1; i <= N; i++)
    {
        if(vis[i] == false && i != checkedCity)
        {
            vis[i] = true;
            dfs(i);
            ans++;
        }
    }
    return ans - 1;
}

int main()
{ 
    // freopen("data.txt", "r", stdin);
    scanf("%d%d%d", &N, &M, &K);
    int u, v;
    for (int i = 0; i < M; i++)
    {   // 讀取道路總數, 保存道路之間關係
        scanf("%d%d", &u, &v);
        G[u][v] = G[v][u] = 1;
    }
    int city;
    for (int i = 0; i < K; i++)
    {
        scanf("%d", &checkedCity);
        printf("%d\n", dfsTravel(city));
    }

    return 0;
}