L2-013. 红色警报 (并查集)

战争中保持各个城市间的连通性非常重要。本题要求你编写一个报警程序，当失去一个城市导致国家被分裂为多个无法连通的区域时，就发出红色警报。注意：若该国本来就不完全连通，是分裂的k个区域，而失去一个城市并不改变其他城市之间的连通性，则不要发出警报。

输入格式：

输入在第一行给出两个整数N（0 < N <=500）和M（<=5000），分别为城市个数（于是默认城市从0到N-1编号）和连接两城市的通路条数。随后M行，每行给出一条通路所连接的两个城市的编号，其间以1个空格分隔。在城市信息之后给出被攻占的信息，即一个正整数K和随后的K个被攻占的城市的编号。

注意：输入保证给出的被攻占的城市编号都是合法的且无重复，但并不保证给出的通路没有重复。

输出格式：

对每个被攻占的城市，如果它会改变整个国家的连通性，则输出“Red Alert: City k is lost!”，其中k是该城市的编号；否则只输出“City k is lost.”即可。如果该国失去了最后一个城市，则增加一行输出“Game Over.”。

输入样例：

5 4
0 1
1 3
3 0
0 4
5
1 2 0 4 3

输出样例：

City 1 is lost.
City 2 is lost.
Red Alert: City 0 is lost!
City 4 is lost.
City 3 is lost.
Game Over.

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 500 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
bool vis[maxn];
vector<int> G[maxn];
int p[maxn];
inline int Find(int x){ return x == p[x] ? x : p[x] = Find(p[x]); }
bool cnt[maxn];

int main(){
  scanf("%d %d", &n, &m);
  for(int i = 0; i < n; ++i)  p[i] = i;
  for(int i = 0; i < m; ++i){
    int u, v;
    scanf("%d %d", &u, &v);
    G[u].push_back(v);
    int x = Find(u);
    int y = Find(v);
    if(x != y)  p[y] = x;
  }
  scanf("%d", &m);
  for(int j = 0; j < m; ++j){
    int t;
    scanf("%d", &t);
    vis[t] = 1;
    int tt = Find(t);
    memset(cnt, 0, sizeof cnt);
    for(int i = 0; i < n; ++i)  if(!vis[i] && Find(i) == tt)  cnt[i] = 1;
    for(int i = 0; i < n; ++i)  if(cnt[i])  p[i] = i;
    for(int i = 0; i < n; ++i)  if(cnt[i]){
      int x = Find(i);
      for(int k = 0; k < G[i].size(); ++k) if(cnt[G[i][k]]){
        int y = Find(G[i][k]);
        if(x != y)  p[y] = x;
      }
    }
    set<int> sets;
    for(int i = 0; i < n; ++i)  if(cnt[i])  sets.insert(Find(i));
    if(sets.size() < 2)  printf("City %d is lost.\n", t);
    else printf("Red Alert: City %d is lost!\n", t);
  }
  if(m == n)  puts("Game Over.");
  return 0;
}