bzoj 1015 并查集，离线

1015: [JSOI2008]星球大战starwar

题意：n个点，m条双向边，k个询问。每次删掉一个点，问连通块个数。

tags：一开始按顺序做，发现搞不出来。。离线做，从后往前，把删除变为添加。

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define rep(i,a,b) for (int i=a;i<=b;i++)
#define per(i,b,a) for (int i=b;i>=a;i--)
#define mes(a,b)  memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
typedef long long ll;
const int N = 4e5+10;

struct Edge{int to, next;}e[N<<1];
int n, m, k, fa[N], head[N], ans[N], q[N], tot, sum;
bool flag[N];
void Addedge(int u, int v) {e[++tot].to=v, e[tot].next=head[u], head[u]=tot;}
int Find(int x) {return x==fa[x] ? x : fa[x]=Find(fa[x]);}
void add(int u)
{
    flag[u]=0, sum++;
    for(int i=head[u]; i; i=e[i].next) {
        int v=e[i].to;
        if(flag[v]==0) {
            int fau=Find(u), fav=Find(v);
            if(fau!=fav) fa[fav]=fau, sum--;
        }
    }
}
int main()
{
    scanf("%d %d", &n, &m);
    rep(i,0,n-1) fa[i]=i;
    int u, v;
    rep(i,1,m) {
        scanf("%d %d", &u, &v);
        Addedge(u, v); Addedge(v, u);
    }
    scanf("%d", &k);
    rep(i,1,k) scanf("%d", &q[i]), flag[q[i]]=1;
    rep(i,0,n-1) if(flag[i]==0) add(i);
    ans[k+1]=sum;
    per(i,k,1) add(q[i]), ans[i]=sum;
    rep(i,1,k+1) printf("%d\n", ans[i]);

    return 0;
}