CCCC L2-013. 红色警报 连通分量

题解：将问题转化成连通分量。每次失去一座城市，切断其所有的边，算一次现在的连通分量。若增量大于1，则发出警报。

　　　　至于如何算连通分量，直接用tarjan模板

坑：//我昨天晚上半夜敲的模板，把一个算所有环中最短环的tarjan模板 直接贴上去了，一直不过，还XJB改了会儿233（现在90行的代码逛逛B站随便改改就ac了2333）

 #define _CRT_SECURE_NO_WARNINGS
 #include<iostream>
 #include<vector>
 #include<queue>
 #include<map>
 #include<cmath>
 #include<stack>
 #include<string.h>
 #include<set>
 using namespace std;
 const int maxn =  + ;
 vector<int> E[maxn];
 int dfn[maxn], low[maxn], tot, n, ans =, vis[maxn];
 stack<int> S;
 void init() {
     for (int i = ; i < maxn; i++) vis[i] = dfn[i] = low[i] = ;
     tot = ;
     ans = ;
 }
 void tarjan(int x) {
     low[x] = dfn[x] = ++tot;
     S.push(x); vis[x] = ;
     for (int i = ; i < E[x].size(); i++) {
         int v = E[x][i];
         if (!dfn[v]) {
             tarjan(v);
             low[x] = min(low[x], low[v]);

         }
         else if (vis[x]) {
             low[x] = min(low[x], dfn[v]);
         }
     }
     if (low[x] == dfn[x]) {
         //int cnt = 0;
         while () {
             int now = S.top();
             S.pop();
             vis[x] = ;
             //cnt++;
             if (now == x)break;
         }
         ans++;
         //if (cnt > 1)ans = min(ans, cnt);
     }
 }
 set<int> st[maxn];
 int main() {
     int m;
     cin >> n >> m;
     for (int i = ; i < n; i++)E[i].push_back(i);
     for (int i = ; i <= m; i++) {
         int x, y;
         scanf("%d%d", &x, &y);//判重？？没用
         if (st[x].count(y) == ) { E[x].push_back(y); st[x].insert(y); }
         if (st[y].count(x) == ) { E[y].push_back(x); st[y].insert(x); }
         //E[y].push_back(x);
     }
     //n500m5000
     for (int i = ; i <= n; i++) {
         if (!dfn[i])tarjan(i);
     }
     int last = ans;
     int q; cin >> q;

     for (int j = ; j <= q; j++) {
         int x; cin >> x;
         int ok = ;
         for (int i = ; i < E[x].size(); i++) {
             int v = E[x][i];
 //            for (auto t : E[v]) if (t == x)t = -1;//强行去除反向边。
             for (vector<int>::iterator it = E[v].begin(); it != E[v].end(); it++)if (*it == x) { E[v].erase(it); break; }
         }

         E[x].clear();
         E[x].push_back(x);
         init();
         //cout << last<< endl;
         for (int i = ; i <= n; i++) {
             if (!dfn[i])tarjan(i);
         }
         if (ans > last + ) ok = ;
         last = ans;
         if (ok)printf("Red Alert: City %d is lost!\n", x);
         else printf("City %d is lost.\n", x);
     }
     //cout << ans << endl;
     if (q == n)cout << "Game Over.";
     system("pause");
 }