题意：有三种类型的路

类型 1. 从 U 到 V  能走INF个人 能躲 d 个人

类型 2. 从 U 到 V 能走INF 个人 不能 躲人

类型 3. 从 U 到 V 能走INF 个人 要用该路时需要花费一定费用

要求能躲最多人 且花费最少

类型3 的路最多有 12 条

所以我们用二进制 枚举 类型3

（也有递归来枚举，dfs心太累）

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <cmath>
using namespace std;
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#include <set>
#include <map>
#include <time.h>;
#define cler(arr, val)    memset(arr, val, sizeof(arr))
#define FOR(i,a,b)  for(int i=a;i<=b;i++)
#define IN   freopen ("in.txt" , "r" , stdin);
#define OUT  freopen ("out.txt" , "w" , stdout);
typedef long long  LL;
const int MAXN = 200;
const int MAXM = 100000;
const int INF = 0x3f3f3f3f;
//const int INF = 922222;
const int mod = 1000000007;
struct Edge
{int to,next,cap,flow;
} edge[MAXM]; //注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],cur[MAXN];
void init()
{tol = 0;memset(head,-1,sizeof (head));
}
void addedge (int u,int v,int w,int rw = 0)
{edge[tol].to = v;edge[tol].cap = w;edge[tol].flow = 0;edge[tol].next = head[u];head[u] = tol++;edge[tol].to = u;edge[tol].cap = rw;edge[tol].flow = 0;edge[tol].next = head[v];head[v] = tol++;
}
int Q[MAXN];
void BFS(int start,int end)
{memset(dep,-1,sizeof(dep));memset(gap,0,sizeof(gap));gap[0] = 1;int front = 0, rear = 0;dep[end] = 0;Q[rear++] = end;while(front != rear){int u = Q[front++];for(int i = head[u]; i !=  -1; i = edge[i].next){int v = edge[i]. to;if(dep[v] != -1)continue;Q[rear++] = v;dep[v] = dep[u] + 1;gap[dep[v]]++;}}
}
int S[MAXN];
int sap(int start,int end, int N)
{BFS(start,end);memcpy(cur,head,sizeof(head));int top = 0;int u = start;int ans = 0;int i;while(dep[start] < N){if(u == end){int Min = INF;int inser;for( i = 0; i < top; i++){if(Min > edge[S[i]].cap - edge[S[i]].flow){Min = edge[S[i]].cap - edge[S[i]].flow;inser = i;}}for( i = 0; i < top; i++){edge[S[i]]. flow += Min;edge[S[i]^1].flow -= Min;}ans += Min;top = inser;u = edge[S[top]^1].to;continue;}bool flag =  false;int v;for( i = cur[u]; i != -1; i = edge[i]. next){v = edge[i]. to;if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u]){flag =  true;cur[u] = i;break;}}if(flag){S[top++] = cur[u];u = v;continue;}int Min = N;for( i = head[u]; i !=  -1; i = edge[i].next){if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min){Min = dep[edge[i].to];cur[u] = i;}}gap[dep[u]]--;if(!gap[dep[u]]) return ans;dep[u] = Min + 1;gap[dep[u]]++;if(u != start)u = edge[S[--top]^1].to;}return ans;
}
int u[MAXM],v[MAXM],d[MAXM],p[MAXM],man[MAXN];
int main()
{
#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);
#endifint n,m;while(scanf("%d%d",&n,&m)!=EOF){int star=0,end=n+1,mansum,num3=0,num1=0;for(int i=1;i<=n;i++){scanf("%d",&man[i]);if(d!=0)mansum+=man[i];}for(int i=0;i<m;i++){scanf("%d%d%d%d",&u[i],&v[i],&d[i],&p[i]);if(p[i]<0&&d[i]!=0)num1++;else if(p[i]>0)num3++;}if(!num1)//优化了 下快了很多{puts("Poor Heaven Empire");continue;}int maxflow=0,cost=0;for(int k=0;k<(1<<num3);k++)//枚举类型3 {init();int x=end,cc=0,cos=0,ans;for(int i=1;i<=n;i++)addedge(star,i,man[i]);for(int i=0;i<m;i++){if(p[i]<0){addedge(u[i],v[i],INF);addedge(u[i],end,d[i]);}else if(p[i]==0){addedge(u[i],v[i],INF);}else if(p[i]>0){if(k&(1<<cc)){addedge(u[i],v[i],INF);cos+=d[i];}else addedge(u[i],v[i],1);cc++;}}ans=sap(star,end,n+2);if(ans>maxflow)maxflow=ans,cost=cos;else if(ans==maxflow&&cost>cos)cost=cos;}if(maxflow==0)puts("Poor Heaven Empire");else printf("%d %d\n",maxflow,cost);}return 0;
}