[JSOI 2008]星球大战starwar

Description

题库链接

给你一张 \(n\) 点， \(m\) 条边的无向图，每次摧毁一个点，问你剩下几个联通块。

\(1\leq n\leq 2m,1\leq m\leq 200000\)

Solution

删点不好操作，我们考虑倒序，变为加点。加边时，只考虑没删除的点间的连边，并查集维护。

是一道喜闻乐见的大水题。

Code

//It is made by Awson on 2018.2.27
#include <bits/stdc++.h>
#define LL long long
#define dob complex<double>
#define Abs(a) ((a) < 0 ? (-(a)) : (a))
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
#define writeln(x) (write(x), putchar('\n'))
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int N = 400000;
void read(int &x) {
    char ch; bool flag = 0;
    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
    x *= 1-2*flag;
}
void print(int x) {if (x > 9) print(x/10); putchar(x%10+48); }
void write(int x) {if (x < 0) putchar('-'); print(Abs(x)); }

int n, m, u, v, t, a[N+5], ans[N+5], flag[N+5];
struct tt {int to, next, cost; }edge[(N<<2)+5];
int path[N+5], top, cnt, fa[N+5];

void add(int u, int v) {edge[++top].to = v, edge[top].next = path[u], path[u] = top; }
int find(int o) {return fa[o] ? fa[o] = find(fa[o]) : o; }
void work() {
    read(n), read(m); cnt = n;
    for (int i = 1; i <= m; i++) {
    read(u), read(v); ++u, ++v; add(u, v), add(v, u);
    }
    read(t); for (int i = 1; i <= t; i++) read(a[i]), ++a[i], flag[a[i]] = 1; cnt -= t;
    for (int u = 1; u <= n; u++)
    if (flag[u] == 0)
        for (int j = path[u]; j; j = edge[j].next) {
        if (flag[edge[j].to] == 1) continue;
        if (find(u)^find(edge[j].to)) fa[find(u)] = find(edge[j].to), --cnt;
        }
    ans[t+1] = cnt;
    for (int i = t; i >= 1; i--) {
    flag[a[i]] = 0; int u = a[i]; ++cnt;
    for (int j = path[u]; j; j = edge[j].next)
        if (flag[edge[j].to] == 0)
        if (find(u)^find(edge[j].to)) fa[find(u)] = find(edge[j].to), --cnt;
    ans[i] = cnt;
    }
    for (int i = 1; i <= t+1; i++) writeln(ans[i]);
}
int main() {
    work(); return 0;
}