P1231 教辅的组成

传送门：https://www.luogu.org/problemnew/show/P1231

这是一道很不错的网络流入门题，关键在于如何建图。

首先，我们将练习册和源点连一条边权为1的边，然后若书 i 和练习册 j 可以配套，就将连一条从练习册 j 到书 i 边，当然边权还是1。同理，答案和书也是如此，最后再将答案和汇点连一条边权为1的边。

但是这么写还是会有点问题，因为经过一本书的路径可能与很多条，书就被使用了多次，显然不符合题意。这时候我们可以将书 i 拆成书 i1 和 i2，i1 和练习册连边，i2 和答案连边，这样就保证没一本书之用过一次了。

建好图后跑最大流就行了。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<stack>
#include<queue>
#include<vector>
using namespace std;
#define enter printf("
")
#define space printf(" ")
#define Mem(a) memset(a, 0, sizeof(a))
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e4 + 5;
inline ll read()
{
    ll ans = 0;
    char ch = getchar(), last = ' ';
    while(!isdigit(ch)) {last = ch; ch = getchar();}
    while(isdigit(ch))
    {
        ans = ans * 10 + ch - '0'; ch = getchar();
    }
    if(last == '-') ans = -ans;
    return ans;
}
inline void write(ll x)
{
    if(x < 0) x = -x, putchar('-');
    if(x >= 10) write(x / 10);
    putchar(x % 10 + '0');
}

int t, n1, n2, n3;

struct Edge
{
    int from, to, cap, flow;
};
vector<Edge> edges;
vector<int> G[maxn << 2];
void addEdge(int from, int to)
{
    edges.push_back((Edge){from, to, 1, 0});
    edges.push_back((Edge){to, from, 0, 0});
    int sz = edges.size();
    G[from].push_back(sz - 2);
    G[to].push_back(sz - 1);
}

int dis[maxn << 2];
bool vis[maxn << 2];
bool bfs()
{
    Mem(vis);
    queue<int> q;
    q.push(0); vis[0] = 1;
    dis[0] = 0;
    while(!q.empty())
    {
        int now = q.front(); q.pop();
        for(int i = 0; i < (int)G[now].size(); ++i)
        {
            Edge& e = edges[G[now][i]];
            if(!vis[e.to] && e.cap > e.flow)
            {
                vis[e.to] = 1;
                dis[e.to] = dis[now] + 1;
                q.push(e.to);
            }
        }
    }
    return vis[t];
}
int cur[maxn << 2];
int dfs(int now, int a)
{
    if(now == t || !a) return a;
    int flow = 0, f;
    for(int& i = cur[now]; i < (int)G[now].size(); ++i)
    {
        Edge& e = edges[G[now][i]];
        if(dis[e.to] == dis[now] + 1 && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0)
        {
            e.flow += f;
            edges[G[now][i] ^ 1].flow -= f;        
            flow += f; a -= f;
            if(!a) break;    
        }
    }
    return flow;
}

int maxflow()
{
    int flow = 0;
    while(bfs())
    {
        Mem(cur);
        flow += dfs(0, INF);
    }
    return flow;
}

//bool vis2[maxn];

int main()
{
    n1 = read(); n2 = read(); n3 = read();         
    t = (n1 << 1)+ n2 + n3 + 1;                //为了防止编号重复
/*按注释掉的写法会WA掉，因为如果一本书多次输入，那么这个书
就多次拆点，这一本书就可以使用多次了 */ 
/*    int m = read();                            
    while(m--)
    {
        int x = read(), y = read();        
        addEdge(0, y + (n1 << 1));
        addEdge(y + (n1 << 1), x);
        addEdge(x, x + n1);
        vis2[x] = 1;
    }
    m = read();
    while(m--)
    {
        int x = read(), y = read();
        if(!vis2[x]) addEdge(x, x + n1), vis2[x] = 1;
        addEdge(x + n1, y + (n1 << 1) + n2);
        addEdge(y + (n1 << 1) + n2, t);
    }*/
    int m = read();
    while(m--)
    {
        int x = read(), y = read();
        addEdge(y + (n1 << 1), x);
    }
    m = read();
    while(m--)
    {
        int x = read(), y = read();
        addEdge(x + n1, y + (n1 << 1) + n2);
    }
    for(int i = 1; i <= n1; ++i) addEdge(i, i + n1);
    for(int i = 1; i <= n2; ++i) addEdge(0, i + (n1 << 1));
    for(int i = 1; i <= n3; ++i) addEdge(i + (n1 << 1) + n2, t);
    write(maxflow()); enter;
}