题做多的话不难想到可能是以行列作为二分图两个点集，然后网络流相关
具体怎么弄呢
我们可以用求补集的思想，假设有解
我们先把棋盘能放的地方放满士兵，然后我们尽量的把士兵拿走
并且要满足行和列的要求，
说到这方法就很明显了
ans=n*m-障碍数-最大流
我们用x[i]表示棋盘放满后第i行最多能移开的士兵数
y[i]表示棋盘放满后第i列最多能移开的士兵数
然后建图就更明显了，不多说
无解预先判断一下即可
const inf=;
type node=record
next,point,flow:longint;
end; var edge:array[..] of node;
he,li,cur,p,numh,pre,h,d:array[..] of longint;
a:array[..,..] of boolean;
k,i,j,n,m,x,y,len,t:longint; function min(a,b:longint):longint;
begin
if a>b then exit(b) else exit(a);
end; procedure add(x,y,f:longint);
begin
inc(len);
edge[len].point:=y;
edge[len].flow:=f;
edge[len].next:=p[x];
p[x]:=len;
end; function sap:longint;
var u,i,j,neck,tmp,q:longint;
begin
neck:=inf;
u:=;
numh[]:=t+;
h[]:=;
sap:=;
for i:= to t do
cur[i]:=p[i];
while h[]<t+ do
begin
d[u]:=neck;
i:=cur[u];
while i<>- do
begin
j:=edge[i].point;
if (edge[i].flow>) and (h[u]=h[j]+) then
begin
neck:=min(neck,edge[i].flow);
pre[j]:=u;
cur[u]:=i;
u:=j;
if u=t then
begin
sap:=sap+neck;
while u<> do
begin
u:=pre[u];
j:=cur[u];
dec(edge[j].flow,neck);
inc(edge[j xor ].flow,neck);
end;
neck:=inf;
end;
break;
end;
i:=edge[i].next;
end;
if i=- then
begin
dec(numh[h[u]]);
if numh[h[u]]= then exit;
tmp:=t;
i:=p[u];
q:=-;
while i<>- do
begin
j:=edge[i].point;
if edge[i].flow> then
if h[j]<tmp then
begin
tmp:=h[j];
q:=i;
end;
i:=edge[i].next;
end;
h[u]:=tmp+;
cur[u]:=q;
inc(numh[h[u]]);
if u<> then
begin
u:=pre[u];
neck:=d[u];
end;
end;
end;
end; begin
len:=-;
fillchar(p,sizeof(p),);
readln(n,m,k);
t:=n+m+;
for i:= to n do
read(he[i]);
for i:= to m do
read(li[i]);
for i:= to k do
begin
readln(x,y);
a[x,y]:=true;
inc(he[x]);
inc(li[y]);
end;
for i:= to n do
begin
if he[i]>m then
begin
writeln('JIONG!');
halt;
end;
add(,i,m-he[i]);
add(i,,);
end;
for i:= to m do
begin
if li[i]>n then
begin
writeln('JIONG!');
halt;
end;
add(i+n,t,n-li[i]);
add(t,i+n,);
end;
for i:= to n do
for j:= to m do
if not a[i,j] then
begin
add(i,j+n,);
add(j+n,i,);
end;
writeln(n*m-k-sap);
end.