[luoguP2045] 方格取数加强版（最小费用最大流）

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水题

——代码

 #include <queue>
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #define N 51
 #define M 100001
 #define INF 1e9
 #define min(x, y) ((x) < (y) ? (x) : (y))

 int n, k, s, t, cnt, tot, sum;
 int head[M], to[M], val[M], cost[M], next[M], dis[M], pre[M], a[N][N], b[N][N];
 bool vis[M];

 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 add2(int x, int y, int z, int c)
 {
     to[cnt] = y;
     val[cnt] = z;
     cost[cnt] = c;
     next[cnt] = head[x];
     head[x] = cnt++;
 }

 inline void add(int x, int y, int z, int c)
 {
     add2(x, y, z, c);
     add2(y, x, , -c);
 }

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

 int main()
 {
     int i, j, x, d;
     n = read();
     k = read();
     s = , t = (n * n << ) + ;
     memset(head, -, sizeof(head));
     for(i = ; i <= n; i++)
         for(j = ; j <= n; j++)
         {
             x = read(), b[i][j] = ++tot;
             add(b[i][j], b[i][j] + n * n, , -x);
             add(b[i][j], b[i][j] + n * n, INF, );
         }
     for(i = ; i <= n; i++)
         for(j = ; j <= n; j++)
         {
             if(i < n) add(b[i][j] + n * n, b[i + ][j], INF, );
             if(j < n) add(b[i][j] + n * n, b[i][j + ], INF, );
         }
     add(s, , k, );
     add(n * n << , t, k, );
     while(spfa())
     {
         d = INF;
         for(i = pre[t]; i ^ -; i = pre[to[i ^ ]]) d = min(d, val[i]);
         for(i = pre[t]; i ^ -; i = pre[to[i ^ ]])
         {
             val[i] -= d;
             val[i ^ ] += d;
         }
         sum += dis[t] * d;
     }
     printf("%d\n", -sum);
     return ;
 }