题目描述：

给你一些房间，和门，告诉你门从哪一侧可以开。初始时，所有的门都是开的。有一些房间有坏蛋。你在第i号房间，问你最少关多少道门，使所有坏蛋都不能到达i号房间。

解题报告：

原点s和所有有坏蛋的房间连接，容量无穷.

a房间可以到b房间，从a房间可以开，连接a，b容量无穷，连接b，a，容量1

求原点到i点的最小割。

如果是无穷，无解，否则最小割就是答案。

代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define Max 0x1fffffff
#define size 100
struct edge{int from, to, val, next;}e[140000];
int v[size], que[size], dis[size], cnt, cur[size];
void insert(int from, int to, int va)
{
    e[cnt].from = from, e[cnt].to = to; e[cnt].val = va;
    e[cnt].next = v[from];v[from] = cnt++;
    e[cnt].from = to, e[cnt].to = from; e[cnt].val = 0;
    e[cnt].next = v[to];v[to] = cnt++;
}
bool bfs(int n, int s, int t)
{
    int head, tail, id;
    head = tail = 0; que[tail++] = s;
    memset(dis, -1, sizeof(int) * n);dis[s] = 0;
    while(head < tail) // bfs,得到顶点i的距s的最短距离dis[i]
        for(id = v[que[head++]]; id != -1; id = e[id].next)
            if (e[id].val > 0 && dis[e[id].to] == -1)
            {
                dis[e[id].to] = dis[e[id].from] + 1;
                que[tail++] = e[id].to;
                if (e[id].to == t) return true;
            }
    return false;
}

int Dinic(int n, int s, int t)
{
    int maxflow = 0, tmp, i;
    while(bfs(n, s, t))
    {
        int u = s, tail = 0;
        for(i = 0; i < n; i++) cur[i] = v[i];
        while(cur[s] != -1)
            if (u != t && cur[u] != -1 && e[cur[u]].val > 0 && dis[u] != -1 && dis[u] + 1 == dis[e[cur[u]].to])
            {que[tail++] = cur[u]; u = e[cur[u]].to;}
            else if (u == t)
            {
                for(tmp = Max, i = tail - 1; i >= 0; i--) tmp = min(tmp, e[que[i]].val);
                for(maxflow += tmp, i = tail - 1; i >= 0; i--)
                {
                    e[que[i]].val -= tmp;
                    e[que[i] ^ 1].val += tmp;
                    if (e[que[i]].val == 0) tail = i;
                }
                u = e[que[tail]].from;
            }
            else
            {
                while(tail > 0 && u != s && cur[u] == -1) u = e[que[--tail]].from;
                cur[u] = e[cur[u]].next;
            }
    }
    return maxflow;
}
int ca, n, t;
char str[5];
int main()
{
    scanf("%d", &ca);
    while(ca--)
    {
        scanf("%d%d", &n, &t);
        int s = n;
        memset(v, -1, sizeof(int) * (s + 1)); cnt = 0;
        for(int i = 0; i < n; i++)
        {
            int num;
            scanf("%s%d", str, &num);
            if (str[0] == 'I') insert(s, i, Max);
            for(int j = 0; j < num; j++)
            {
                int tmp;scanf("%d", &tmp);
                insert(i, tmp, Max); insert(tmp, i, 1);
            }
        }
        int ans = Dinic(s + 1, s, t);
        if (ans >= Max) printf("PANIC ROOM BREACH\n");
        else printf("%d\n", ans);
    }
    return 0;
}