题意：给定n*m的图，每个士兵可以占领当前行和列，第i行至少要R[i]个士兵占领，第j列至少要C[j]个士兵占领，部分网格无法占领，求占领所用最少士兵数，若无解则输出orz
士兵的贡献情况有1(只有效占领行/列)，2（既占领行又占领列）
用最大流跑出贡献为2的士兵的个数，然后把所有要求相减处理就得出贡献为1的士兵个数
非法方案用极端情况和要求行列士兵数去考虑
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define rrep(i,j,k) for(register int i=j;i>=k;i--)
#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])
#define iin(a) scanf("%d",&a)
#define lin(a) scanf("%lld",&a)
#define din(a) scanf("%lf",&a)
#define s0(a) scanf("%s",a)
#define s1(a) scanf("%s",a+1)
#define print(a) printf("%lld",(ll)a)
#define enter putchar('\n')
#define blank putchar(' ')
#define println(a) printf("%lld\n",(ll)a)
#define IOS ios::sync_with_stdio(0)
using namespace std;
const int MAXN = 1e6+11;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-7;
typedef long long ll;
ll read(){
ll x=0,f=1;register char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int to[MAXN<<1],nxt[MAXN<<1],cap[MAXN<<1],flow[MAXN<<1];
int head[MAXN],tot;
void init(){
memset(head,-1,sizeof head);
tot=0;
}
void add(int u,int v,int w){
to[tot]=v;
nxt[tot]=head[u];
cap[tot]=w;
flow[tot]=0;
head[u]=tot++;
swap(u,v);
to[tot]=v;
nxt[tot]=head[u];
cap[tot]=0;
flow[tot]=0;
head[u]=tot++;
}
int n,m,s,t;
int dis[MAXN],pre[MAXN],cur[MAXN],gap[MAXN];
bool vis[MAXN];
struct QUEUE{
int que[MAXN];
int front,rear;
void init(){front=rear=0;}
void push(int u){que[rear++]=u;}
int pop(){return que[front++];}
bool empty(){return front==rear;}
}que;
void bfs(){
memset(vis,0,sizeof vis);
que.init();
que.push(t);
vis[t]=1;dis[t]=0;
while(que.empty()^1){
int u = que.pop();
for(int i = head[u]; ~i; i = nxt[i]){
register int v=to[i],c=cap[i^1],f=flow[i^1];
if(!vis[v]&&c>f){
vis[v]=1;
dis[v]=dis[u]+1;
que.push(v);
}
}
}
}
int aug(){
int u=t,ans=INF;
while(u!=s){
ans=min(ans,cap[pre[u]]-flow[pre[u]]);
u=to[pre[u]^1];
}
u=t;
while(u!=s){
flow[pre[u]]+=ans;
flow[pre[u]^1]-=ans;
u=to[pre[u]^1];
}
return ans;
}
int isap(){
int ans=0;
bfs();
memset(gap,0,sizeof gap);
memcpy(cur,head,sizeof head);
for(int i = 1; i <= n; i++) gap[dis[i]]++;
int u = s;
while(dis[s]<n){
if(u==t){
ans+=aug();
u=s;
}
bool ok=0;
for(int i = cur[u]; ~i; i = nxt[i]){
int v=to[i],c=cap[i],f=flow[i];
if(c>f&&dis[u]==dis[v]+1){
ok=1;
pre[v]=i;
cur[u]=i;
u=v;
break;
}
}
if(!ok){
int mn=n-1;
for(int i = head[u]; ~i; i = nxt[i]){
int v=to[i],c=cap[i],f=flow[i];
if(c>f) mn=min(mn,dis[v]);
}
if(--gap[dis[u]]==0) break;
dis[u]=mn+1;gap[dis[u]]++;cur[u]=head[u];
if(u!=s) u=to[pre[u]^1];
}
}
return ans;
}
int r,c,k;
int R[233],C[333];
int block[233][333];
int haver[233],havec[233];
int main(){
while(cin>>r>>c>>k){
init(); int cnt=0;
memset(block,0,sizeof block);
memset(haver,0,sizeof haver);
memset(havec,0,sizeof havec);
rep(i,1,r) R[i]=read();
rep(i,1,c) C[i]=read();
rep(i,1,r) cnt+=R[i];
rep(i,1,c) cnt+=C[i];
rep(i,1,k){
int a=read();
int b=read();
block[a][b]=1;
}
s=r+c+1;t=s+1;n=t;
rep(i,1,r) add(s,i,R[i]);
rep(i,1,c) add(i+r,t,C[i]);
rep(i,1,r) rep(j,1,c){
if(!block[i][j]){
haver[i]++;
havec[j]++;
add(i,j+r,1);
}
}
bool flag=0;
rep(i,1,r) if(haver[i]<R[i]) flag=1;
rep(i,1,c) if(havec[i]<C[i]) flag=1;
if(flag){
printf("JIONG!\n");
continue;
}
ll ans=isap();
ll tmp=cnt-2*ans;
println(ans+tmp);
}
return 0;
}