费用流，两个点间连费用为负价值容量为1的边，再连费用为零容量为INF的边，建立S，T，分别向起点终点连边，跑最小费用流，对答案取反即可

水

# include 
     
      # define RG register# define IL inline# define Fill(a, b) memset(a, b, sizeof(a))using namespace std;typedef long long ll;const int _(1010), __(1e6 + 10), INF(2e9);IL ll Read(){    char c = '%'; ll x = 0, z = 1;    for(; c > '9' || c < '0'; c = getchar()) if(c == '-') z = -1;    for(; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';    return x * z;}int a, b, p, q, id[20][20];int cnt, fst[_], w[__], to[__], nxt[__], dis[_], vis[_], S, T, cost[__], pe[_], pv[_], max_flow, max_cost;queue 
      
        Q;IL void Add(RG int u, RG int v, RG int f, RG int co){    cost[cnt] = co; w[cnt] = f; to[cnt] = v; nxt[cnt] = fst[u]; fst[u] = cnt++;    cost[cnt] = -co; w[cnt] = 0; to[cnt] = u; nxt[cnt] = fst[v]; fst[v] = cnt++;}IL bool Bfs(){    Q.push(S); Fill(dis, 127); dis[S] = 0; vis[S] = 1;    while(!Q.empty()){        RG int u = Q.front(); Q.pop();        for(RG int e = fst[u]; e != -1; e = nxt[e]){            if(!w[e] || dis[to[e]] <= dis[u] + cost[e]) continue;            dis[to[e]] = dis[u] + cost[e];            pe[to[e]] = e; pv[to[e]] = u;            if(!vis[to[e]]) vis[to[e]] = 1, Q.push(to[e]);        }        vis[u] = 0;    }    if(dis[T] >= dis[T + 1]) return 0;    RG int ret = INF;    for(RG int u = T; u != S; u = pv[u]) ret = min(ret, w[pe[u]]);    for(RG int u = T; u != S; u = pv[u]) w[pe[u]] -= ret, w[pe[u] ^ 1] += ret;    max_cost -= ret * dis[T]; max_flow += ret;    return 1;}int main(RG int argc, RG char *argv[]){    Fill(fst, -1);    a = Read(); b = Read(); p = Read(); q = Read();    for(RG int i = 0; i <= p; ++i)        for(RG int j = 0; j <= q; ++j)            id[i][j] = ++cnt;    T = cnt + 1; cnt = 0;    for(RG int i = 0; i <= p; ++i)        for(RG int j = 1, v; j <= q; ++j)            v = Read(), Add(id[i][j - 1], id[i][j], 1, -v), Add(id[i][j - 1], id[i][j], INF, 0);    for(RG int j = 0; j <= q; ++j)        for(RG int i = 1, v; i <= p; ++i)            v = Read(), Add(id[i - 1][j], id[i][j], 1, -v), Add(id[i - 1][j], id[i][j], INF, 0);    for(RG int i = 1, k, x, y; i <= a; ++i) k = Read(), x = Read(), y = Read(), Add(S, id[x][y], k, 0);    for(RG int i = 1, k, x, y; i <= b; ++i) k = Read(), x = Read(), y = Read(), Add(id[x][y], T, k, 0);    while(Bfs()); printf("%d\n", max_cost);    return 0;}