题目链接：https://www.patest.cn/contests/gplt/L2-013
求联通块的个数，因为时限给的很宽，我们可以每一次都直接重新并查集建图处理，再来统计连通块的个数。
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 505;
int set[maxn],n;
struct Road
{
int u,v;
}r[maxn*10];
int cmp(Road a,Road b)
{
return a.u < b.u;
}
int find(int x)
{
int r = x;
while(r != set[r])
r = set[r];
return r;
}
void merge(int a,int b)
{
int x = find(a);
int y = find(b);
if(x != y) set[x] = y;
}
void init()
{
for(int i=0; i<n; i++) set[i] = i;
}
int main()
{
int m;
scanf("%d%d", &n,&m);
init();
for(int i=0; i<m; i++)
{
scanf("%d%d", &r[i].u,&r[i].v);
merge(r[i].u,r[i].v);
}
int sum = 0;
for(int i=0; i<n; i++)
if(set[i] == i) sum++;
int q;
scanf("%d", &q);
for(int j=0; j<q; j++)
{
int tmp;
scanf("%d", &tmp);
init();
int num = 0;
for(int i=0; i<m; i++)
{
if(r[i].u == tmp || r[i].v == tmp) r[i].u = maxn-1,num++;
else merge(r[i].u,r[i].v);
}
sort(r,r+m,cmp);
m -= num;
num = 0;
for(int i=0; i<n; i++)
if(set[i] == i) num++;
if(num - sum > 1) printf("Red Alert: ");
printf("City %d is lost",tmp);
if(num - sum > 1) printf("!\n");
else printf(".\n");
sum = num;
if(j == n-1) printf("Game Over.\n");
}
}