战争中的城市

https://www.acwing.com/problem/content/1487/

并查集维护连通块的写法

#include <iostream>
#include <cstring>

using namespace std;

const int N = 1010, M = 500010;

int n, m, k;
int p[N];

struct Edge
{
    int a, b;
}e[M];

int find(int x)
{
    if (p[x] != x) p[x] = find(p[x]);
    return p[x];
}

int main()
{
    scanf("%d%d%d", &n, &m, &k);

    for (int i = 0; i < m; i ++ ) scanf("%d%d", &e[i].a, &e[i].b);

    while (k -- )
    {
        int x;
        scanf("%d", &x);

        for (int i = 1; i <= n; i ++ ) p[i] = i;

        int cnt = n - 1;
        for (int i = 0; i < m; i ++ )
        {
            int a = e[i].a, b = e[i].b;
            if (a != x && b != x)
            {
                int pa = find(a), pb = find(b);
                if (pa != pb)
                {
                    p[pa] = pb;
                    cnt -- ;
                }
            }
        }

        printf("%d\n", cnt - 1);
    }

    return 0;
}

dfs维护连通块的写法

#include <iostream>
#include <vector>
#include <memory.h>


using namespace std;


const int N = 1010;

vector<int> graph[N];
int vis[N];
int lost,ans;

int n,m,k;
int a,b;

void dfs(int x){
    vis[x]=1;
    int size = graph[x].size();
    for(int i =0;i < size;i++){
        int p = graph[x][i];
        if(vis[p]==0 && p!=lost){
            dfs(p);   
        }
    }
}


void solve(){
    ans =0;
    memset(vis,0,sizeof vis);
    for(int i = 1;i<=n;i++){
        if(i!=lost && vis[i] ==0){
            ans++; dfs(i);    
        }
    }

    printf("%d\n",ans-1);
    return ;
}

int main(){
    scanf("%d%d%d",&n,&m,&k);

    for(int i = 1; i <= m;i++){
        scanf("%d%d",&a,&b);
        graph[a].push_back(b);
        graph[b].push_back(a);
    }

    for(int i =1;i<=k;i++){
        scanf("%d",&lost);
        solve();
    }

    return 0;
}