PAT(甲级)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

简析：给出一个无向图，问去掉一个点后，其他的至少需要多少线能够连通。主要是运用dfs，求出去掉一个点后所剩下的块数，块数减一就是所得的解。

#include<iostream>
#include<vector>
#include<cstring>
using namespace std;
const int N = 1111;
vector<int> G[N];
bool vis[N];

int currentPoint;

void dfs(int v)
{
    if(v == currentPoint)
        return;
    vis[v] = true;
    for (int i = 0; i < G[v].size();i++)
    {
        if(vis[G[v][i]] == false)
            dfs(G[v][i]);
    }
}
int n, m, k;
int main()
{
    cin >> n >> m >> k;
    for (int i = 0; i < m;i++)
    {
        int a, b;
        cin >> a >> b;
        G[a].push_back(b);
        G[b].push_back(a);
    }
    for (int q = 0; q < k;q++)
    {
        cin >> currentPoint;
        memset(vis, false, sizeof(vis));
        int block = 0;
        for (int i = 1; i <= n;i++)
        {
            if(i != currentPoint && vis[i] == false)
            {
                dfs(i);
                block++;
            }
        }
        cout << (block - 1) << '\n';
    }
    return 0;
}