非常好的网络流
每个顾客分别用一个结点来表示。
对于每个猪圈的第一个顾客，从源点向他连一条边，容量就是该猪圈里的猪的初始数量
对于每个猪圈，假设有n个顾客打开过它，则对所有整数i∈[1, n)，从该猪圈的第i个顾客向第i + 1个顾客连一条边，容量为无穷。
从各个顾客到汇点各有一条边，容量是各个顾客能买的数量上限。
其实很好理解
1 const inf=100000007; 2 type node=record 3 from,point,flow,next:longint; 4 end; 5 var p,s,w,h,numh,cur,pre:array[0..1010] of longint; 6 v:array[0..1010] of boolean; 7 a:array[0..1010,0..1010] of longint; 8 edge:array[0..1000010] of node; 9 ans,z,len,t,i,j,k,x,y,n,m:longint; 10 11 procedure add(x,y,f:longint); 12 begin 13 inc(len); 14 edge[len].from:=x; 15 edge[len].point:=y; 16 edge[len].flow:=f; 17 edge[len].next:=p[x]; 18 p[x]:=len; 19 end; 20 21 procedure sap; 22 var q,u,i,j,flow,neck,tmp:longint; 23 begin 24 fillchar(numh,sizeof(numh),0); 25 fillchar(h,sizeof(h),0); 26 fillchar(pre,sizeof(pre),255); 27 numh[0]:=t+1; 28 u:=0; 29 while h[0]<t+1 do 30 begin 31 if u=t then 32 begin 33 neck:=0; 34 flow:=inf; 35 i:=0; 36 j:=cur[i]; 37 while i<>t do 38 begin 39 if flow>edge[j].flow then 40 begin 41 flow:=edge[j].flow; 42 neck:=i; 43 end; 44 i:=edge[j].point; 45 j:=cur[i]; 46 end; 47 i:=0; 48 j:=cur[i]; 49 while i<>t do 50 begin 51 dec(edge[j].flow,flow); 52 inc(edge[j xor 1].flow,flow); 53 i:=edge[j].point; 54 j:=cur[i]; 55 end; 56 ans:=ans+flow; 57 u:=neck; 58 end; 59 q:=-1; 60 i:=p[u]; 61 while i<>-1 do 62 begin 63 x:=edge[i].point; 64 if (edge[i].flow>0) and (h[u]=h[x]+1) then 65 begin 66 q:=i; 67 break; 68 end; 69 i:=edge[i].next; 70 end; 71 if q<>-1 then 72 begin 73 cur[u]:=q; 74 pre[x]:=u; 75 u:=x; 76 end 77 else begin 78 dec(numh[h[u]]); 79 if numh[h[u]]=0 then break; 80 tmp:=t+1; 81 i:=p[u]; 82 while i<>-1 do 83 begin 84 x:=edge[i].point; 85 if (edge[i].flow>0) then tmp:=min(tmp,h[x]); 86 i:=edge[i].next; 87 end; 88 h[u]:=tmp+1; 89 inc(numh[h[u]]); 90 if u<>0 then u:=pre[u]; 91 end; 92 end; 93 end; 94 95 96 begin 97 readln(m,n); 98 fillchar(p,sizeof(p),255); 99 len:=-1; 100 t:=n+1; 101 for i:=1 to m do 102 read(w[i]); 103 for i:=1 to n do 104 begin 105 read(y); 106 z:=0; 107 for j:=1 to y do 108 begin 109 read(x); 110 if v[x]=false then 111 begin 112 v[x]:=true; 113 z:=z+w[x]; //如果多条边连接源点和顾客，那么合并 114 end; 115 inc(s[x]); 116 a[x,s[x]]:=i; 117 end; 118 if z<>0 then 119 begin 120 add(0,i,z); 121 add(i,0,0); 122 end; 123 readln(x); 124 add(i,t,x); 125 add(t,i,0); 126 end; 127 for i:=1 to m do 128 for j:=1 to s[i]-1 do 129 for k:=j+1 to s[i] do 130 begin 131 add(a[i,j],a[i,k],inf); 132 add(a[i,k],a[i,j],0); 133 end; 134 sap; 135 writeln(ans); 136 end.