点击链接PAT甲级-AC全解汇总

题目：
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 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
1
2
3
4

Sample Output:

1
0
0
1
2
3

题意：
输入 结点数N 链接数M 需要检查的个数K
然后M行输入M对 代表链接，在之后一行K个数代表要查询的结点
题目意思是，去掉当前节点后剩下需要填几条线可以连成一个通路，那么很容易想到计算连通分量的个数，输出连通分量个数-1就行

我的代码：

#include
using namespace std;

int N,M,K;
bool citys[1010][1010]={0};
bool marked[1010]={0};

void DFS(int start){
    marked[start]=true;
    for(int i=1;i<=N;i++)
    {
        if(citys[start][i]==false||marked[i])continue;
        else DFS(i);
    }
}

int main()
{
    memset(citys,0,sizeof(citys));
    cin>>N>>M>>K;
    for(int i=0;i<M;i++)
    {
        int a,b;
        cin>>a>>b;
        citys[a][b]=true;
        citys[b][a]=true;
    }

    for(int i=0;i<K;i++)
    {
//        memset(marked,0,sizeof(marked));
        fill(marked,marked+1010,false);
        int t,cnt=0;
        cin>>t;
        marked[t]=true;
        for(int j=1;j<=N;j++)
        {
            if(marked[j])continue;
            DFS(j);
            cnt++;
        }
        cout<<cnt-1<<endl;
    }

    return 0;
}

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注意： 用memset的话会超时，用fill不会