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("\n")
 #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-;
 const int maxn = 1e4 + ;
 inline ll read()
 {
     ll ans = ;
     char ch = getchar(), last = ' ';
     while(!isdigit(ch)) {last = ch; ch = getchar();}
     while(isdigit(ch))
     {
         ans = ans *  + ch - ''; ch = getchar();
     }
     if(last == '-') ans = -ans;
     return ans;
 }
 inline void write(ll x)
 {
     if(x < ) x = -x, putchar('-');
     if(x >= ) write(x / );
     putchar(x %  + '');
 }

 int t, n1, n2, n3;

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

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

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

 //bool vis2[maxn];

 int main()
 {
     n1 = read(); n2 = read(); n3 = read();
     t = (n1 << )+ n2 + n3 + ;                //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 << ), x);
     }
     m = read();
     while(m--)
     {
         int x = read(), y = read();
         addEdge(x + n1, y + (n1 << ) + n2);
     }
     for(int i = ; i <= n1; ++i) addEdge(i, i + n1);
     for(int i = ; i <= n2; ++i) addEdge(, i + (n1 << ));
     for(int i = ; i <= n3; ++i) addEdge(i + (n1 << ) + n2, t);
     write(maxflow()); enter;
 }