POJ 1149 PIGS 学会构图，最大流

这道题知道怎么构图之后就不难了，然而我一开始并没有想出来怎么构图，之后看了题解。大概理解了图是怎么建的，按照题解的套路写完A掉后也没有仔细去思考到底是怎样如此简洁地构图的。但是学会了一种思路，这个也是从那篇题解中看到的：

在面对网络流问题时，如果一时想不出很好的构图方法，不如先构造一个最直观，或者说最“硬来”的模型，然后再用合并节点和边的方法来简直化这个模型。经过简化以后，好的构图思路自然就会涌现出来了。这是解决网络流问题的一个好方法。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#define INF (1<<25)
using namespace std;

int n, m, beg[1005], maxbuy[105], firbuy[1005];
vector <int> pighou[1005];

struct Edge {int u, v, flow, cap;};

struct Dinic
{
    int s, t, e, d[111], cur[111];
    vector <int> G[111];
    Edge E[22222];

    void addedge (int u, int v, int cap)
    {
        G[u].push_back(e), G[v].push_back(e+1);
        E[e].u = u, E[e].v = v, E[e++].cap = cap;
        E[e].u = v, E[e].v = u, E[e++].cap = 0;
    }

    bool BFS ()
    {
        memset (d, -1, sizeof d);
        queue <int> q;
        q.push(s); d[s] = 0;
        while (!q.empty())
        {
            int u = q.front(); q.pop();
            for (int i = G[u].size()-1; i >= 0; i--)
            {
                Edge e0 = E[G[u][i]];
                if (d[e0.v] < 0 && e0.flow < e0.cap)
                    d[e0.v] = d[u]+1, q.push(e0.v);
            }
        }
        return d[t] >= 0;
    }

    int DFS (int u, int flow)
    {
        if (u == t || !flow) return flow;

        int f1 = 0, f2;
        for (int &i = cur[u]; i >= 0; i--)
        {
            Edge &e1 = E[G[u][i]];
            Edge &e2 = E[G[u][i]^1];
            if (d[e1.v] != d[u] + 1) continue;
            f2 = DFS(e1.v,min(flow,e1.cap-e1.flow));
            if (f2 > 0) 
                f1 += f2, e1.flow += f2,
                flow -= f2, e2.flow -= f2;
            if (!flow) break;
        }
        return f1;
    }

    void work ()
    {
        int mxf = 0;
        while (BFS())
        {
            for (int i = s; i <= t; i++) cur[i] = G[i].size()-1;
            mxf += DFS(s,1<<30);
        }
        printf ("%d\n", mxf);
    }
}Solve;

void ReadinData ()
{
    scanf ("%d %d", &m, &n);
    for (int i = 1; i <= m; i++) scanf ("%d", beg+i);
    for (int i = 1; i <= n; i++)
    {
        int num, tmp; scanf ("%d", &num);
        for (int j = 1; j <= num; j++)
        {
            scanf ("%d", &tmp);
            pighou[tmp].push_back(i);
            if (!firbuy[tmp]) firbuy[tmp] = i;
        }
        scanf ("%d", maxbuy+i);
    }
}

void BuildGraph ()
{
    Solve.s = 0, Solve.t = n+1;
    for (int i = 1; i <= m; i++) Solve.addedge(Solve.s,firbuy[i],beg[i]);
    for (int i = 1; i <= n; i++) Solve.addedge(i,Solve.t,maxbuy[i]);
    for (int i = 1; i <= m; i++)
    for (int j = pighou[i].size()-1; j > 0; j--)
        Solve.addedge(pighou[i][j-1],pighou[i][j],INF);
}

int main ()
{
    ReadinData();
    BuildGraph();
    Solve.work();
    return 0;
}