[luoguP2045] 方格取数加强版（最小费用最大流）

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水题

——代码

  1 #include <queue>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <iostream>
  5 #define N 51
  6 #define M 100001
  7 #define INF 1e9
  8 #define min(x, y) ((x) < (y) ? (x) : (y))
  9
 10 int n, k, s, t, cnt, tot, sum;
 11 int head[M], to[M], val[M], cost[M], next[M], dis[M], pre[M], a[N][N], b[N][N];
 12 bool vis[M];
 13
 14 inline int read()
 15 {
 16     int x = 0, f = 1;
 17     char ch = getchar();
 18     for(; !isdigit(ch); ch = getchar()) if(ch == ‘-‘) f = -1;
 19     for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - ‘0‘;
 20     return x * f;
 21 }
 22
 23 inline void add2(int x, int y, int z, int c)
 24 {
 25     to[cnt] = y;
 26     val[cnt] = z;
 27     cost[cnt] = c;
 28     next[cnt] = head[x];
 29     head[x] = cnt++;
 30 }
 31
 32 inline void add(int x, int y, int z, int c)
 33 {
 34     add2(x, y, z, c);
 35     add2(y, x, 0, -c);
 36 }
 37
 38 inline bool spfa()
 39 {
 40     int i, u, v;
 41     std::queue <int> q;
 42     memset(vis, 0, sizeof(vis));
 43     memset(pre, -1, sizeof(pre));
 44     memset(dis, 127 / 3, sizeof(dis));
 45     q.push(s);
 46     dis[s] = 0;
 47     while(!q.empty())
 48     {
 49         u = q.front(), q.pop();
 50         vis[u] = 0;
 51         for(i = head[u]; i ^ -1; i = next[i])
 52         {
 53             v = to[i];
 54             if(val[i] && dis[v] > dis[u] + cost[i])
 55             {
 56                 dis[v] = dis[u] + cost[i];
 57                 pre[v] = i;
 58                 if(!vis[v])
 59                 {
 60                     q.push(v);
 61                     vis[v] = 1;
 62                 }
 63             }
 64         }
 65     }
 66     return pre[t] ^ -1;
 67 }
 68
 69 int main()
 70 {
 71     int i, j, x, d;
 72     n = read();
 73     k = read();
 74     s = 0, t = (n * n << 1) + 1;
 75     memset(head, -1, sizeof(head));
 76     for(i = 1; i <= n; i++)
 77         for(j = 1; j <= n; j++)
 78         {
 79             x = read(), b[i][j] = ++tot;
 80             add(b[i][j], b[i][j] + n * n, 1, -x);
 81             add(b[i][j], b[i][j] + n * n, INF, 0);
 82         }
 83     for(i = 1; i <= n; i++)
 84         for(j = 1; j <= n; j++)
 85         {
 86             if(i < n) add(b[i][j] + n * n, b[i + 1][j], INF, 0);
 87             if(j < n) add(b[i][j] + n * n, b[i][j + 1], INF, 0);
 88         }
 89     add(s, 1, k, 0);
 90     add(n * n << 1, t, k, 0);
 91     while(spfa())
 92     {
 93         d = INF;
 94         for(i = pre[t]; i ^ -1; i = pre[to[i ^ 1]]) d = min(d, val[i]);
 95         for(i = pre[t]; i ^ -1; i = pre[to[i ^ 1]])
 96         {
 97             val[i] -= d;
 98             val[i ^ 1] += d;
 99         }
100         sum += dis[t] * d;
101     }
102     printf("%d\n", -sum);
103     return 0;
104 }