[HNOI2013]切糕

最小割

---

让相邻的最小割上的点层数差小于等于d，也就是大于d时S，T仍然能相通
那么可以从下面d层向上面相邻的点连容量为INF的边
表示相邻的最小割上的点层数差大于d时，还能走这条INF边回来，流其它的路到T，这样就强制要小于等于d了

---

# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
# define Copy(a, b) memcpy(a, b, sizeof(a))
# define ID(a, b) n * (a - 1) + b
using namespace std;
typedef long long ll;
const int _(2e5 + 10), __(2e6 + 10), INF(2147483647);

IL ll Read(){
    RG char c = getchar(); RG ll x = 0, z = 1;
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int p, q, r, v[60][60][60], d, id[60][60][60], num;
int w[__], fst[_], nxt[__], to[__], cnt, S, T, lev[_], cur[_], max_flow;
queue <int> Q;

IL void Add(RG int u, RG int v, RG int f){
    w[cnt] = f; to[cnt] = v; nxt[cnt] = fst[u]; fst[u] = cnt++;
    w[cnt] = 0; to[cnt] = u; nxt[cnt] = fst[v]; fst[v] = cnt++;
}

IL int Dfs(RG int u, RG int maxf){
    if(u == T) return maxf;
    RG int ret = 0;
    for(RG int &e = cur[u]; e != -1; e = nxt[e]){
        if(lev[to[e]] != lev[u] + 1 || !w[e]) continue;
        RG int f = Dfs(to[e], min(w[e], maxf - ret));
        ret += f; w[e ^ 1] += f; w[e] -= f;
        if(ret == maxf) break;
    }
    if(!ret) lev[u] = 0;
    return ret;
}

IL bool Bfs(){
    Fill(lev, 0); lev[S] = 1; Q.push(S);
    while(!Q.empty()){
        RG int u = Q.front(); Q.pop();
        for(RG int e = fst[u]; e != -1; e = nxt[e]){
            if(lev[to[e]] || !w[e]) continue;
            lev[to[e]] = lev[u] + 1;
            Q.push(to[e]);
        }
    }
    return lev[T];
}

int main(RG int argc, RG char* argv[]){
    Fill(fst, -1); p = Read(); q = Read(); r = Read(); d = Read();
    for(RG int i = 1; i <= r; ++i)
        for(RG int j = 1; j <= p; ++j)
            for(RG int k = 1; k <= q; ++k)
                id[i][j][k] = ++num, v[i][j][k] = Read();
    for(RG int j = 1; j <= p; ++j)
        for(RG int k = 1; k <= q; ++k)
            id[r + 1][j][k] = ++num;
    T = num + 1;
    for(RG int j = 1; j <= p; ++j)
        for(RG int k = 1; k <= q; ++k)
            Add(S, id[1][j][k], INF), Add(id[r + 1][j][k], T, INF);
    for(RG int i = 1; i <= r; ++i)
        for(RG int j = 1; j <= p; ++j)
            for(RG int k = 1; k <= q; ++k){
                Add(id[i][j][k], id[i + 1][j][k], v[i][j][k]);
                if(i - d){
                    if(j) Add(id[i][j][k], id[i - d][j - 1][k], INF);
                    if(k) Add(id[i][j][k], id[i - d][j][k - 1], INF);
                    if(j < p) Add(id[i][j][k], id[i - d][j + 1][k], INF);
                    if(k < q) Add(id[i][j][k], id[i - d][j][k + 1], INF);
                }
            }
    while(Bfs()) Copy(cur, fst), max_flow += Dfs(S, INF);
    printf("%d
", max_flow);
    return 0;
}