uva 10330【 Power Transmission】

第一道网络流的题目，我是看这个讲解的http://www.nocow.cn/index.php/%E7%BD%91%E7%BB%9C%E6%B5%81Edmonds-Karp 算法

主要是顶点有值限制，所以可以将一个点分成两个点a和a'，它们两个之间的容量就是顶点的值

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

const int maxn = 300;
const int inf = 0x7fffffff;
int g[maxn][maxn];
int pre[maxn];
int n,m;

int maxflow(int end)
{
    int ans = 0;
    while(true)
    {
        queue<int> q;
        q.push(0);
        memset(pre,-1,sizeof(pre));
        while(!q.empty())
        {
            int u = q.front();
            q.pop();
            for(int v = 1;v <= end&&pre[end] < 0;v ++)
            {
                if(g[u][v] > 0 && pre[v] == -1)
                {
                    pre[v] = u;
                    q.push(v);
                }
                if(pre[end] >= 0)
                    break;
            }
        }
        if(pre[end] < 0)
            break;
        int min = -1;
        for(int v = end;v > 0;v = pre[v])
        {
            if(min == -1 || min > g[pre[v]][v])
                min = g[pre[v]][v];
        }
        int v;
        for(int t = pre[v = end];t >= 0;v = t,t = pre[v])
        {
            g[t][v] -= min;
            g[v][t] += min;
        }
        ans += min;

    }

    return ans;
}

int main()
{
    while(scanf("%d",&n) == 1)
    {
        memset(g,0,sizeof(g));
        int a,b,c;
        for(int i = 1;i <= n;i ++)
        {
            scanf("%d",&a);
            g[i][i+n] = a;
        }
        scanf("%d",&m);
        for(int i = 0;i < m;i ++)
        {
            scanf("%d%d%d",&a,&b,&c);
            g[a+n][b] = c;
        }
        int B,D;
        scanf("%d%d",&B,&D);
        for(int i = 0;i < B;i ++)
        {
            scanf("%d",&a);
            g[0][a] = inf;
        }

        for(int i = 0;i < D;i ++)
        {
            scanf("%d",&a);
            g[a+n][2*n + 1] =  inf;
        }

        printf("%d\n",maxflow(2*n+1));
    }

    return 0;
}