最大流 这题有很多起点和终点 在取2个点（0和n+1） 作为唯一的起点和终点
此外每个点也有容量限制 建图时每条边上的容量为这条边和2个端的容量的最小值 然后EK就行
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int MAX = 110;
int num[MAX];
int a[MAX];
int cap[MAX][MAX];
int flow[MAX][MAX];
int p[MAX];
int n,m,b,d;
int f;
void EK()
{
queue <int> q;
memset(flow,0,sizeof(flow));
f = 0;
while(1)
{
memset(a,0,sizeof(a));
a[0] = 999999999;
q.push(0);
while(!q.empty())
{
int u = q.front();
q.pop();
for(int v = 0; v <= n+1; v++)
{
if(!a[v] && cap[u][v] > flow[u][v])
{
p[v] = u;
q.push(v);
a[v] = min(a[u],cap[u][v] - flow[u][v]);
}
}
}
if(a[n+1] == 0)
break;
for(int u = n+1; u; u = p[u])
{
flow[p[u]][u] += a[n+1];
flow[u][p[u]] -= a[n+1];
}
f += a[n+1];
}
}
int main()
{
int i,x,y,z;
while(scanf("%d",&n)!=EOF)
{
memset(cap,0,sizeof(cap));
for(i = 1;i <= n; i++)
scanf("%d",&num[i]);
scanf("%d",&m);
while(m--)
{
scanf("%d %d %d",&x,&y,&z);
cap[x][y] = z;
cap[x][y] = min(cap[x][y],min(num[x],num[y]));
}
scanf("%d %d",&b,&d);
while(b--)
{
scanf("%d",&x);
cap[0][x] = num[x];
}
while(d--)
{
scanf("%d",&x);
cap[x][n+1] = num[x];
}
EK();
printf("%d\n",f);
}
return 0;
}