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;
}