题意：
给几个发电站，给几个消耗站，再给几个转发点。
发电站只发电，消耗站只消耗电，转发点只是转发电，再给各个传送线的传电能力。
问你消耗站能取得的最多电是多少。
思路：增加1个超级源点，和超级汇点。。把所给的发电站都和超级源点相连，把所给的消耗战都和超级汇点相连。。用EK求最大流。
模板有几个地方要注意。
1：start是编号最前的点，last是编号最后的点
模板默许last是m，根据需要要把m都改成编号最后的点的号码
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<queue>
#include<stdlib.h>
#define inf 0x3f3f3f3f
using namespace std;
int map[1000][1000], path[1000], flow[1000], n, m,nc,np,start;
int last;
queue<int>que;
int bfs()
{
int i, t;
while (que.size())que.pop();
memset(path, ⑴, sizeof(path));
path[start] = 0;
flow[start] = inf;
que.push(start);
while (que.size())
{
int t = que.front();
que.pop();
if (t == last) break;
for (i = 0; i <= n+1; i++)
{
if (i != start && path[i] == ⑴ && map[t][i])
{
flow[i] = min(flow[t], map[t][i]);
path[i] = t;
que.push(i);
}
}
}
if (path[last] == ⑴)return ⑴;
else return flow[n+1];
}
int liu()
{
int maxflow = 0, temp, now, pre;
while ((temp = bfs()) != ⑴)
{
maxflow += temp;
now = last;
while (now != start)
{
pre = path[now];
map[pre][now] -= temp;
map[now][pre] += temp;
now = pre;
}
}
return maxflow;
}
char temp[20];
int main()
{
int i, j, k, from, to, cost,u,v,z;
while (~scanf("%d%d%d%d", &n,&np,&nc,&m))
{
memset(map, 0, sizeof(map));
for (i = 0; i < m; i++)
{
scanf("%s", temp); sscanf(temp, "(%d,%d)%d", &u, &v, &z);
map[u+1][v+1] += z;
}
for (i = 0; i < np; i++)
{
scanf("%s", temp); sscanf(temp, "(%d)%d", &u, &z);
map[0][u+ 1] += z;
}
for (i = 0; i < nc; i++)
{
scanf("%s", temp); sscanf(temp, "(%d)%d", &u, &z);
map[u+1][n+1] += z;
}
last=n + 1;
start = 0;
cout << liu() << endl;
}
}