刚学OI ->0 秒，求助

已经做了一天了qwq，

思路就是先离线，然后反过来从最后炸完了回到最初还没被炸的时候。

\kkkkkkkk \kel

然后这里是代码：

#include <cstdio>

const int N = 4e5 + 7, M = 2e5 + 7;
int n, m, k, cnt, ans, first[N], fa[N], time[M], otp[M];//cnt、first是前向星的，time表示第几次炸的是哪一个，ans是统计当前的连通块，otp是输出的答案(也就是记录了所有的ans)
bool damaged[N];//表示是否被炸
struct edges{
    int v, next;
}edge[M<<1];

inline void add_edge(int u, int v){
    edge[++ cnt] = (edges){v, first[u]};
    first[u] = cnt ++;
    edge[cnt] = (edges){u, first[v]};
    first[v] = cnt;
}
inline int find(int x){return fa[x] == x?x:fa[x] = find(fa[x]);}
inline bool check(int x, int y){return find(x) == find(y);}
inline bool merge(int x, int y){
    if (check(x, y))return false;
    return fa[fa[x]] = fa[y], true;
}

int main(){
    scanf("%d %d", &n, &m);
    for (register int i = 0;i < n;i ++)fa[i] = i;
    for (register int i = 1, x, y;i <= m;add_edge(x, y), i ++)scanf("%d %d", &x, &y);
    scanf("%d", &k);
    ans = n - k;
    for (register int i = 1;i <= k;damaged[time[i]] = true, i ++)scanf("%d", time + i);
    for (register int i = 0;i < n;i ++)
        if (damaged[i])continue;
        else for (register int j = first[i];j;j = edge[j].next)if (!damaged[edge[j].v] && merge(i, edge[j].v))ans --;
    otp[k + 1] = ans;
    for (register int i = k;i;otp[i] = ++ ans, i --)for (register int j = first[time[i]];j;j = edge[j].next)if (!damaged[edge[j].v] && merge(time[i], edge[j].v))ans --;
    for (register int i = 1;i <= k + 1;i ++)printf("%d\n", otp[i]);
    return 0;
}