[luoguP2763] 试题库问题（最大流）

传送门

每个类别和它所有的试题连一条权值为1的边。

增加一个超级源点s，s和每个类别连一条权值为选当前类别数量的边。

增加一个超级汇点t，每个试题和t连一条权值为1的边。

求最大流即可。

——代码

 #include <queue>
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #define min(x, y) ((x) < (y) ? (x) : (y))
 #define N 1100
 #define M 3000001

 int k, n, m, cnt, s, t, sum;
 int num[N], a[N][N];
 int head[N], to[M], next[M], val[M], dis[N], cur[N];

 inline int read()
 {
     int x = , f = ;
     char ch = getchar();
     for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -;
     for(; isdigit(ch); ch = getchar()) x = (x << ) + (x << ) + ch - '';
     return x * f;
 }

 inline void add(int x, int y, int z)
 {
     to[cnt] = y;
     val[cnt] = z;
     next[cnt] = head[x];
     head[x] = cnt++;
 }

 inline bool bfs()
 {
     int i, u, v;
     std::queue <int> q;
     memset(dis, -, sizeof(dis));
     q.push(s);
     dis[s] = ;
     while(!q.empty())
     {
         u = q.front(), q.pop();
         for(i = head[u]; i ^ -; i = next[i])
         {
             v = to[i];
             if(val[i] && dis[v] == -)
             {
                 dis[v] = dis[u] + ;
                 if(v == t) return ;
                 q.push(v);
             }
         }
     }
     return ;
 }

 inline int dfs(int u, int maxflow)
 {
     if(u == t) return maxflow;
     int i, v, d, ret = ;
     for(i = cur[u]; i ^ -; i = next[i])
     {
         v = to[i];
         if(val[i] && dis[v] == dis[u] + )
         {
             d = dfs(v, min(val[i], maxflow - ret));
             ret += d;
             cur[u] = i;
             val[i] -= d;
             val[i ^ ] += d;
             if(ret == maxflow) return ret;
         }
     }
     return ret;
 }

 int main()
 {
     int i, j, p, x;
     k = read();
     n = read();
     s = , t = n + k + ;
     for(i = ; i <= k; i++) m += num[i] = read();
     for(i = ; i <= n; i++)
     {
         p = read();
         for(j = ; j <= p; j++) x = read(), a[x][++a[x][]] = i;
     }
     memset(head, -, sizeof(head));
     for(i = ; i <= n; i++) add(i, t, ), add(t, i, );
     for(i = ; i <= k; i++) add(s, i + n, num[i]), add(i + n, s, );
     for(i = ; i <= k; i++)
         for(j = ; j <= a[i][]; j++)
             add(i + n, a[i][j], ), add(a[i][j], i + n, );
     while(bfs())
     {
         for(i = s; i <= t; i++) cur[i] = head[i];
         sum += dfs(s, 1e9);
     }
     if(sum ^ m)
     {
         puts("No Solution!");
         return ;
     }
     for(i = n + ; i <= n + k; i++)
     {
         printf("%d:", i - n);
         for(j = head[i]; j ^ -; j = next[j])
             if(!val[j] && to[j] ^ s)
                 printf(" %d", to[j]);
         puts("");
     }
     return ;
 }