UVa 10330 Power Transmission / 最大流EK

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;
}