题目链接:点击打开链接
题目大意:有n个人,m个小组。一个人可能在多个小组,只要与0在一个小组的人都会可能被感染,这些人也可能感染别人and so on...现在让你求可能被感染的人数。
解题思路:显然的并查集求解,最后求一下谁和0的父亲节点是一样的就行了
代码:
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<string>
#include<map>
#include<set>
#include<vector>
#define FAST ios::sync_with_stdio(false)
typedef long long ll;
const int mod = (int)1e9 + 7;
const int inf = 0x3f3f3f3f;
const int maxn = (int)3e5 + 5;
using namespace std;
int pre[maxn];
int find(int x){
int r = x;
while(pre[r] != r)
r = pre[r];
int i = x, t;
while(i != r){
t = pre[i];
pre[i] = r;
i = t;
}
return r;
}
void join(int x, int y){
int fx = find(x), fy = find(y);
if(fx != fy)
pre[fx] = fy;
}
int main()
{
int n, m;
while(scanf("%d %d", &n, &m)){
if(n == 0 && m == 0) break;
for(int i = 0; i < n; i++) pre[i] = i;
while(m--){
int x, y; scanf("%d %d", &x, &y);
for(int i = 1; i < x; i++){
int t; scanf("%d", &t);
join(t, y);
}
}
int ans = 0;
for(int i = 0; i < n; i++){
if(find(i) == find(0)) ans++;
}
printf("%d\n", ans);
}
return 0;
}
over