hoj 2715 (费用流 拆点）

http://acm.hit.edu.cn/hoj/problem/view?id=2715

将每个格子 i 拆成两个点 i’, i’’并加边(i’, i’’, 1, -Vi), (i’, i’’, ∞, 0), (s, i’, ∞, 0); 控制只有一次能取到宝物。

对相邻的四个格子 j, Hi > Hj 则加边(i’’, j’, ∞, 0);

若格子 i 在边界上则加边(i’’, t, ∞, 0)。

限制增广次数小于等于 K 求最小费用流即可。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <queue>
 #include <cmath>

 using namespace std;

 const int maxn = ;
 const int maxm = ;
 const int inf = 0x3f3f3f3f;

 struct MCMF
 {
     struct Edge
     {
         int v, c, w, next;
     }p[maxm << ];
     int e, head[maxn], dis[maxn], pre[maxn], cnt[maxn], sumFlow, n;
     bool vis[maxn];
     void init(int nt)
     {
         e = ; n = nt;
         memset(head, -, sizeof(head[]) * (n + ) );
     }
     void addEdge(int u, int v, int c, int w)
     {
         p[e].v = v; p[e].c = c; p[e].w = w; p[e].next = head[u]; head[u] = e++;
         swap(u, v);
         p[e].v = v; p[e].c = ; p[e].w = -w; p[e].next = head[u]; head[u] = e++;
     }
     bool spfa(int S, int T)
     {
         queue <int> q;
         for (int i = ; i <= n; ++i)
             vis[i] = cnt[i] = , pre[i] = -, dis[i] = inf;
         vis[S] = , dis[S] = ;
         q.push(S);
         while (!q.empty())
         {
             int u = q.front(); q.pop();
             vis[u] = ;
             for (int i = head[u]; i + ; i = p[i].next)
             {
                 int v = p[i].v;
                 if (p[i].c && dis[v] > dis[u] + p[i].w)
                 {
                     dis[v] = dis[u] + p[i].w;
                     pre[v] = i;
                     if (!vis[v])
                     {
                         q.push(v);
                         vis[v] = ;
                         if (++cnt[v] > n) return ;
                     }
                 }
             }
         }
         return dis[T] != inf;
     }
     int mcmf(int S, int T, int kt)
     {
         sumFlow = ;
         int minFlow = , minCost = ;
         while (spfa(S, T) && (kt--))
         {
             minFlow = inf + ;
             for (int i = pre[T]; i + ; i = pre[ p[i ^ ].v ])
                 minFlow = min(minFlow, p[i].c);
             sumFlow += minFlow;
             for (int i = pre[T]; i + ; i = pre[ p[i ^ ].v ])
             {
                 p[i].c -= minFlow;
                 p[i ^ ].c += minFlow;
             }
             minCost += dis[T] * minFlow;
         }
         return minCost;
     }
     void build(int nt, int kt)
     {
         int nnt = nt * nt;
         init(nnt *  + );
         int val[][], height[][];
         memset(val, 0x3f, sizeof(val));
         memset(height, 0x3f, sizeof(height));
         for (int i = ; i <= nt; ++i)
             for (int j = ; j <= nt; ++j)
                 scanf("%d", &val[i][j]);
         for (int i = ; i <= nt; ++i)
             for (int j = ; j <= nt; ++j)
                 scanf("%d", &height[i][j]);
         for (int i = ; i <= nt; ++i)
             for (int j = ; j <= nt; ++j)
             {
                 int pos = nt * (i - ) + j;
                 addEdge(, pos, inf, );
                 addEdge(pos, pos + nnt, ,-val[i][j]);
                 addEdge(pos, pos + nnt, inf,);
                 if (i ==  || i == nt || j ==  || j == nt)
                     addEdge(pos + nnt, n, inf, );
                 if (height[i][j] > height[i][j - ])
                     addEdge(pos + nnt, pos - , inf, );
                 if (height[i][j] > height[i][j + ])
                     addEdge(pos + nnt, pos + , inf, );
                 if (height[i][j] > height[i - ][j])
                     addEdge(pos + nnt, pos - nt, inf, );
                 if (height[i][j] > height[i + ][j])
                     addEdge(pos + nnt, pos + nt, inf, );
             }
     }
     void solve(int nt, int kt)
     {
         build(nt, kt);
         printf("%d\n", - mcmf(, n, kt));
     }
 }my;
 int main()
 {
     int tcase, n, k;
     scanf("%d", &tcase);
     while (tcase--)
     {
         scanf("%d%d", &n, &k);
         my.solve(n, k);
     }
     return ;
 }