题目大意

有n个问题，m个人，每个人可以解决一些问题，问最少选多少个人可以解决所有问题

---

解题思路

如果一个人解决的问题被别的人包括，那么可以把这个人丢掉

对于一个问题只能由一个人解决，那么直接选这个人

然后枚举每个问题被谁解决即可

---

代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define N 100
using namespace std;
int n, m, x, y, g, pp, ans, tot, dg[N], uf[N], p[N], head[N], a[N][10];
struct rec
{int to, next;
}e[N*6];
void add(int x, int y)
{e[++tot].to = y;e[tot].next = head[x];head[x] = tot;
}
void dfs(int x, int y)
{if (y >= ans) return;if (x > n){ans = min(ans, y);return;}if (p[x]){dfs(x + 1, y);return;}for (int i = head[x]; i; i = e[i].next)if (!uf[e[i].to]){for (int j = 1; j <= a[e[i].to][0]; ++j)p[a[e[i].to][j]]++;dfs(x + 1, y + 1);for (int j = 1; j <= a[e[i].to][0]; ++j)p[a[e[i].to][j]]--;}return;
}
int main()
{	ans = 60;scanf("%d%d", &n, &m);for(int i = 1; i <= m; ++i){scanf("%d", &x);a[i][0] = x;for (int j = 1; j <= x; ++j){scanf("%d", &a[i][j]);add(a[i][j], i);}sort(a[i] + 1, a[i] + 1 + x);}for (int i = 1; i <= m; ++i)for (int j = 1; j <= m; ++j)//i包括j{if (i == j) continue;pp = 1;x = 1;y = 1;while(x <= a[i][0]){while (a[i][x] == a[j][y] && y <= a[j][0]) y++;if (y > a[j][0]) break;x++;}if (y <= a[j][0]) pp = 0;if (a[i][0] == a[j][0] && i < j) pp = 0;if (pp) uf[j] = 1;}for (int i = 1; i <= m; ++i)if (!uf[i])for (int j = 1; j <= a[i][0]; ++j)dg[a[i][j]]++;for (int i = 1; i <= n; ++i)if (dg[i] == 1 && !p[i])//只能由一个人解决{g++;for (int j = head[i]; j; j = e[j].next)if (!uf[e[j].to]){for (int k = 1; k <= a[e[j].to][0]; ++k)p[a[e[j].to][k]]++;}}dfs(1, g);printf("%d", ans);return 0;
}