题意：输入 n 个城市 m 条边，但是边有三种有向边 a b c d，第一种是 d 是 0，那么就是一条普通的路，可以通过无穷多人，如果 d < 0，那么就是隧道，这个隧道是可以藏 c 个人，当然也是通过无穷多人的，如果 d > 0，那么这是一座桥，第一次可以通过一个人，如果修复的话，就可以通过无穷多人，问你最多藏的人数，还有最少花费。
析：只是这样是不能做的，但是题目说了桥不超过 12 个，说实话这个条件太隐蔽了，就是不想让人发现，可惜的是队友没读出来，我也实在是没想出来怎么做，后来一查题解，知道有这个条件，那么很简单了，枚举桥的每一个状态，是修还是不修，每次跑一次最大流，进去判断，下面说一下怎么建图。
建立一个超级源点 s 和超级汇点 t，对于每个城市，从 s 向每个城市连一条边，容量就是城市人数，然后对于普通的路，那么就直接连接容量无穷大，对于隧道也是直接连接容量无穷大，然后把隧道向汇点 t 连接，容量是可以藏人的数，注意连的隧道的左端点，最后是桥，每次枚举桥的状态，如果是修复，那么就连一条容量无穷大的，如果不修复，那么就连一条容量为 1 的边，然后就OK了。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define be begin()
#define ed end()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
//#define all 1,n,1
#define FOR(i,n,x) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.in", "r", stdin)
#define freopenw freopen("out.out", "w", stdout)
using namespace std; typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 100 + 20;
const int maxm = 1e6 + 10;
const LL mod = 1000000000000000LL;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
inline int readInt(){ int x; scanf("%d", &x); return x; } struct Edge{
int from, to, cap, flow;
}; struct Dinic{
int n, m, s, t;
vector<Edge> edges;
vector<int> G[maxn];
bool vis[maxn];
int d[maxn];
int cur[maxn]; void init(int n){
FOR(i, n, 0) G[i].cl;
edges.cl;
} void addEdge(int from, int to, int cap){
edges.pb((Edge){from, to, cap,0});
edges.pb((Edge){to, from, 0, 0});
m = edges.sz;
G[from].pb(m - 2);
G[to].pb(m - 1);
} bool bfs(){
ms(vis, 0); vis[s] = 1; d[s] = 0;
queue<int> q; q.push(s); while(!q.empty()){
int u = q.front(); q.pop();
for(int i = 0; i < G[u].sz; ++i){
Edge &e = edges[G[u][i]];
if(!vis[e.to] && e.cap > e.flow){
d[e.to] = d[u] + 1;
vis[e.to] = 1;
q.push(e.to);
}
}
}
return vis[t];
} int dfs(int u, int a){
if(u == t || a == 0) return a;
int flow = 0, f;
for(int &i = cur[u]; i < G[u].sz; ++i){
Edge &e = edges[G[u][i]];
if(d[e.to] == d[u] + 1 && (f = dfs(e.to, min(a, e.cap-e.flow))) > 0){
e.flow += f;
edges[G[u][i]^1].flow -= f;
flow += f;
a -= f;
if(a == 0) break;
}
}
return flow;
} int maxFlow(int s, int t){
this->s = s; this->t = t;
int flow = 0;
while(bfs()){ ms(cur, 0); flow += dfs(s, INF); }
return flow;
}
}; Dinic dinic; struct Node{
int u, v, c;
};
vector<Node> bridge; int main(){
while(scanf("%d %d", &n, &m) == 2){
int s = 0, t = n + 1;
dinic.init(t + 5);
bridge.cl;
for(int i = 1; i <= n; ++i) dinic.addEdge(s, i, readInt());
bool ok = false;
for(int i = 1; i <= m; ++i){
int a, b, c, d;
scanf("%d %d %d %d", &a, &b, &c, &d);
if(d == 0) dinic.addEdge(a, b, INF);
else if(d < 0){
dinic.addEdge(a, b, INF);
dinic.addEdge(a, t, c);
ok = true;
}
else bridge.pb((Node){a, b, c});
}
if(!ok){ puts("Poor Heaven Empire"); continue; }
int ans1 = 0, ans2 = 0;
int all = 1<<bridge.sz;
for(int i = 0; i < all; ++i){
int tmp = 0;
for(int j = 0; j < bridge.sz; ++j)
if(i&1<<j){
tmp += bridge[j].c;
dinic.addEdge(bridge[j].u, bridge[j].v, INF);
}
else dinic.addEdge(bridge[j].u, bridge[j].v, 1);
int res = dinic.maxFlow(s, t);
if(res > ans1){
ans1 = res;
ans2 = tmp;
}
else if(res == ans1 && ans2 > tmp) ans2 = tmp;
for(int j = 0; j < bridge.sz; ++j){
dinic.edges.pop_back();
dinic.G[bridge[j].u].pop_back();
dinic.G[bridge[j].v].pop_back();
}
for(int i = 0; i < dinic.edges.sz; ++i)
dinic.edges[i].flow = 0;
}
if(ans1 == 0) puts("Poor Heaven Empire");
else printf("%d %d\n", ans1, ans2);
}
return 0;
}