1 #include <stdio.h>
  2 #include <stdlib.h>
  3 #define M 200001
  4 #define maxx 2000
  5 #define Max(a, b) a > b ? a : b
  6 #define Min(a, b) a < b ? a : b
  7 #define inf (1 << 29)
  8 struct T
  9 {
 10     int u, v, val, next, cost;
 11 } e[M];
 12 int th;
 13 int visit[M], pre[M], dis[M], que[M], g[M], pos[M], map[100][100];
 14 void add(int u, int v, int val, int cost)
 15 {
 16     e[th].u = u, e[th].v = v, e[th].val = val, e[th].cost = cost;
 17     e[th].next = g[u], g[u] = th++;
 18     e[th].u = v, e[th].v = u, e[th].val = 0, e[th].cost = -cost;
 19     e[th].next = g[v], g[v] = th++;
 20 }
 21 int spfa(int s, int t, int n)
 22 {
 23     int i, u, v, k;
 24     for (i = 0; i <= n; i++)
 25     {
 26         pre[i] = -1, visit[i] = 0;
 27     }
 28     int head, tail;
 29     head = tail = 0;
 30     for (i = 0; i <= n; i++) dis[i] = -1;//求最小费用时这里得改成正无穷
 31     que[tail++] = s, pre[s] = s, dis[s] = 0, visit[s] = 1;
 32     while (head != tail)
 33     {
 34         int u = que[head++];
 35         visit[u] = 0;
 36         for (k = g[u]; k != -1; k = e[k].next)
 37         {
 38             v = e[k].v;
 39             if (e[k].val > 0 && dis[u] + e[k].cost > dis[v])//最大费用，求最小费用时这里的改符号
 40             {
 41                 dis[v] = dis[u] + e[k].cost;
 42                 pre[v] = u;
 43                 pos[v] = k;
 44                 if (!visit[v])
 45                 {
 46                     visit[v] = 1;
 47                     que[tail++] = v;
 48                 }
 49             }
 50         }
 51     }
 52     if (pre[t] != -1 && dis[t] > -1)//求最小费用时这里改成dit[t] < 正无穷
 53     {
 54         return 1;
 55     }
 56     return 0;
 57 }
 58 int MinCostFlow(int s, int t, int n)
 59 {
 60     if (s == t)
 61     {
 62         //这里具体情况具体分析，如果有附加s跟t就不用，如果用数据中的点作为s跟t就得考虑
 63         //直接返回某个值。否则spfa里的队列会死循环。
 64     }
 65     int flow = 0, cost = 0;
 66     while (spfa(s, t, n))
 67     {
 68         int u, min = inf;
 69         for (u = t; u != s; u = pre[u])
 70         {
 71             min = Min(min, e[pos[u]].val);
 72         }
 73         flow += min;
 74         cost += dis[t] * min;
 75         for (u = t; u != s; u = pre[u])
 76         {
 77             e[pos[u]].val -= min;
 78             e[pos[u]^1].val += min;
 79         }
 80     }
 81     return cost;
 82 }
 83 int main()
 84 {
 85     int n, m, i, j, k, s, t, u, v, cost, ans, x, num, tt;
 86     while (scanf("%d%d", &n, &m) - EOF)
 87     {
 88         for (i = 0, th = 0; i < n * n * 3 + 2; i++)
 89         {
 90             g[i] = -1;
 91         }
 92         for (i = 0; i < n; i++)
 93         {
 94             for (j = 0; j < n; j++)
 95             {
 96                 scanf("%d", &map[i][j]);
 97             }
 98         }
 99         num = n * n;
100         for (i = 0; i < n; i++)
101         {
102             for (j = 0; j < n; j++)
103             {
104                 x = i * n + j;
105                 add(x, x + num, 1, map[i][j]);
106                 add(x, x + num, inf, 0);
107                 if (i + 1 < n)
108                 {
109                     add(x + num, x + n, inf, 0);
110                 }
111                 if (j + 1 < n)
112                 {
113                     add(x + num, x + 1, inf, 0);
114                 }
115             }
116         }
117         s = 2 * n * n, t = s + 1, n = t + 1;
118         add(s, 0, m, 0);
119         add(s - 1, t, m, 0);
120         ans = MinCostFlow(s, t, t + 1);
121         printf("%d\n", ans);
122     }
123     return 0;
124 }