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 <bits/stdc++.h>
# 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 <int> 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
", max_cost);
return 0;
}