[PTA]L2-013 红色警报

粗略看了一下网上其他题解，都是暴力跑的，也就是对于每个询问重建并查集，这样其实时间复杂度是$O(n^2logn)$的，十分不优秀。

其实有更好的解法，就是时间倒流法，倒序处理每个询问，每个把删去一个点删边改成加上一个点加边，一遍并查集即可。 

每次判断是否合并了两个以上的连通块。

注意一条边可用当且仅当两端的点都存在。

#include <bits/stdc++.h>
#define Mid ((l + r) >> 1)
#define lson (rt << 1)
#define rson (rt << 1 | 1)
using namespace std;
int read(){
    char c; int num, f = 1;
    while(c = getchar(),!isdigit(c)) if(c == '-') f = -1; num = c - '0';
    while(c = getchar(), isdigit(c)) num = num * 10 + c - '0';
    return f * num;
}
const int N = 509;
int n, m, pre[N], a[N], f[N], vis[N];
vector<int> ver[N];
int fid(int x) {
    return pre[x] == x ? x : (pre[x] = fid(pre[x]));
}
signed main()
{
    n = read(); m = read();
    for(int i = 1; i <= m; i++) {
        int x = read() + 1, y = read() + 1;
        ver[x].push_back(y);
        ver[y].push_back(x);
    }
    for(int i = 1; i <= n; i++) pre[i] = i;
    int q = read();
    for(int i = 1; i <= n; i++) vis[i] = 1;
    for(int i = 1; i <= q; i++) vis[a[i] = read() + 1] = 0;
    for(int x = 1; x <= n; x++) if(vis[x])
        for(auto j : ver[x]) 
            if(vis[j] && fid(j) != fid(x)) 
                pre[fid(j)] = fid(x);
    for(int i = q; i; i--) {
        int ff = 0;
        for(auto j : ver[a[i]]) {
            if(vis[j] && fid(j) != fid(a[i])) {
                pre[fid(j)] = fid(a[i]);
                ff++;
            }
        }
        vis[a[i]] = 1;
        if(ff > 1) f[i] = 1;
    }
    for(int i = 1; i <= q; i++) {
        if(f[i]) printf("Red Alert: City %d is lost!
", a[i] - 1);
        else printf("City %d is lost.
", a[i] - 1);
    }
    if(q == n) {
        printf("Game Over.
");
    }
    return 0;
}