【bzoj1015】[JSOI2008]星球大战starwar

给定一个无向图，求联通块个数，以及k次每次摧毁一个点后的联通块个数

将边和摧毁的点全记录下来，反着做即可

注意被摧毁的点不能算作联通块

 

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
 
#define MAXN 300010
 
int n,m;
int k;
 
int key[MAXN<<2],head[MAXN<<2],next[MAXN<<2];
int a[MAXN<<2],fa[MAXN<<2],ans[MAXN<<2];
int kk;
 
bool vis[MAXN<<2];
 
void link(int u,int v)
{
    key[kk]=v;
    next[kk]=head[u];
    head[u]=kk++;
}
 
int find(int x)
{
    return x==fa[x] ? x : fa[x]=find(fa[x]);
}
 
int main()
{
    memset(head,-1,sizeof(head));  
    scanf("%d%d",&n,&m);
    for (int i=1;i<=n;i++)
        fa[i]=i;
    for (int i=1;i<=m;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        link(x,y);
        link(y,x);
    }
    scanf("%d",&k);
    for (int i=1;i<=k;i++)
        scanf("%d",&a[i]),vis[a[i]]=true;
    int z=n-k;
    for (int i=0;i<n;i++)
        if (!vis[i])
            for (int j=head[i];~j;j=next[j])
                if (!vis[key[j]])
                    if (find(i)!=find(key[j]))
                        fa[find(i)]=find(key[j]),z--;
    ans[k+1]=z;
    for (int i=k;i>=1;i--)
    {
        vis[a[i]]=false;
        z++;
        for (int j=head[a[i]];~j;j=next[j])
            if (!vis[key[j]])
                if (find(a[i])!=find(key[j]))
                    fa[find(a[i])]=find(key[j]),z--;
        ans[i]=z;
    }
    for (int i=1;i<=k+1;i++)
        printf("%d\n",ans[i]);
    return 0;
}