L2-013 红色警报 （25 分)（并查集）

额，说实话，还是第一次这样用并查集

本质就是记录连通项的个数。如果，轰炸一个城市，产生一个或者不产生，则说明，该城市对其他城市的连接没有直接作用。

那么怎么来数每次的连通项呢？没错！有多少个祖先就有多少个连通项。其实，就是每次都要把边连接起来，当然，被轰炸的城市所在的边用该丢弃！

#include<iostream>
using namespace std;
const int maxn = 5005;
const int maxx = 505;
struct node{
    int u, v;
}edge[maxn];
int fa[maxx];
int visit[maxx];
int n, m, u, v, k;
int c;
//并查集
int find(int x)
{
    if (x == fa[x])return x;
    return fa[x] = find(fa[x]);
}

void unit(int u, int v){
    int u1 = find(u);
    int v1 = find(v);
    fa[u1] = v1;
}

int main(){
    cin >> n >> m;
    for (int i = 0; i < n; ++i)fa[i] = i;
    for (int i = 0; i < m; ++i)
    {
        cin >> u >> v;
        edge[i].u = u;    edge[i].v = v;
        unit(u, v);
    }
    for (int i = 0; i < n; ++i)find(i);
    //记录原来的连通项数
    int num1 = 0, num2;
    for (int i = 0; i < n; ++i)if (fa[i] == i)num1++;

    cin >> k;
    while(k--){
        cin >> c;        num2 = 0;
        visit[c] = 1;
        for (int i = 0; i < n; ++i)fa[i]=i;
        for (int i = 0; i < m; ++i){
            if (visit[edge[i].u] == 1 || visit[edge[i].v] == 1)continue;
            else {
                unit(edge[i].u, edge[i].v);
            }
        }
        
        for (int i = 0; i < n; ++i)find(i);
        //记录现在的通项数
        for (int i = 0; i < n; ++i)if (fa[i] == i)num2++;
        if (num2  == num1+1 || num2 == num1)cout << "City " << c << " is lost." << endl;
        else cout << "Red Alert: City " << c << " is lost!" << endl;
        num1 = num2;
    }
    num1 = 0;
    for (int i = 0; i < n;++i)
    if (visit[i] == 1)num1++;
    if (num1 == n)cout << "Game Over." << endl;

}