L2-013 红色警报

题目大意

主要思路

首先在读入两条道路通路时用并查集计算出有多少个连通块，
然后记录下来，接下来再读入每一个失去的国家，每读入一个计算一下当前连通块的个数，当当前连通块的个数比上一次的连通块个数多一或者不变时，说明没有改变连通性，否则改变连通性发出红色警报

AC代码

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <string>
#include <cmath>
#include <queue>
#include <vector>
#include <map>
#include <sstream>
using namespace std;

#define x first
#define y second
const int N = 510, M = 5010;
int p[N], n, m, k;
bool st[N];
pair<int, int> pa[M];

void init()
{
    for(int i = 0; i < n; i++) p[i] = i;
}

int count()
{
    int cnt = 0;
    for(int i = 0; i < n; i++)
    {
        if(p[i] == i) cnt++;
    }
    return cnt;
}

int find(int x)
{
    if(p[x] != x) p[x] = find(p[x]);
    return p[x];
}

int main(void)
{
    cin >> n >> m;
    init();
    for(int i = 0; i < m; i++)
    {
        int a, b;
        cin >> a >> b;
        pa[i].x = a, pa[i].y = b;
        a = find(a), b = find(b);
        if(a != b)
        {
            p[a] = b;
        }
    }

    int cnt = count();

    cin >> k;
    for(int i = 0; i < k; i++)
    {
        init();
        int t;
        cin >> t;
        st[t] = true;
        for(int i = 0; i < m; i++)
        {
            int a = pa[i].x, b = pa[i].y;
            if(st[a] || st[b]) continue;
            a = find(a), b = find(b);
            if(a != b)
            {
                p[a] = b;
            }
        }
        int cnt1 = count();
        if(cnt1 > cnt + 1) printf("Red Alert: City %d is lost!\n", t);
        else printf("City %d is lost.\n", t);
        cnt = cnt1;
    }
    if(k == n)
    {
        cout << "Game Over." << endl;
    }
    return 0;
}