[HNOI2013]切糕

嘟嘟嘟


这题我看了半天才懂，一直以为这一刀必须横平竖直的，谁知道这题这一道怎么拐都行，只要满足相邻的高度差不大于\(d\)就行。


看数据范围，猜是最小割。
先不考虑限制。对于每一个\((x, y)\)，我们只能从\(r\)个\(f(x, y)\)选一个，问最小的\(\sum v\)。如果每一个点只有两个选择，那自然往源汇点连边；现在变成了多个选择，那就把这些点穿成一串，再向源汇点连边就好啦（这个技巧得记住啊）
现在有了限制。如果选了这个点的其中一个\(f(z)\)，那么相邻点只能选\([z - d, z + d]\)。也就是割了\(z\)那条边后，只有割\([z - d, z + d]\)这个区间的边才能不让图连通。那我们就从\((x, y, z)\)向\((x', y', z - d)\)和\((x', y', z + d)\)连边。这样如果想断掉\((x, y, z)\)的话，\((x', y')\)只能从\([z - d, z + d]\)选。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 45;
const int maxN = 6e5 + 5;
const int maxe = 2e7 + 5;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int p, q, r, d, t;
int a[maxn][maxn][maxn], id[maxn][maxn][maxn], cnt = 0;
struct Edge
{
  int nxt, from, to, cap, flow;
}e[maxe];
int head[maxN], ecnt = -1;
In void addEdge(int x, int y, int w)
{
  e[++ecnt] = (Edge){head[x], x, y, w, 0};
  head[x] = ecnt;
  e[++ecnt] = (Edge){head[y], y, x, 0, 0};
  head[y] = ecnt;
  //printf("%d %d %d\n", x, y, w);
}

int dis[maxN];
In bool bfs()
{
  Mem(dis, 0); dis[0] = 1;
  queue<int> q; q.push(0);
  while(!q.empty())
    {
      int now = q.front(); q.pop();
      for(int i = head[now], v; i != -1; i = e[i].nxt)
	{
	  if(!dis[v = e[i].to] && e[i].cap > e[i].flow)
	    dis[v] = dis[now] + 1, q.push(v);
	}
    }
  return dis[t];
}
int cur[maxN];
In int dfs(int now, int res)
{
  if(now == t || res == 0) return res;
  int flow = 0, f;
  for(int& i = cur[now], v; i != -1; i = e[i].nxt)
    {
      v = e[i].to;
      if(dis[v] == dis[now] + 1 && (f = dfs(v, min(res, e[i].cap - e[i].flow))) > 0)
	{
	  e[i].flow += f; e[i ^ 1].flow -= f;
	  flow += f; res -= f;
	  if(res == 0) break;
	}
    }
  return flow;
}
In int minCut()
{
  int flow = 0;
  while(bfs())
    {
      memcpy(cur, head, sizeof(head));
      flow += dfs(0, INF);
    }
  return flow;
}

const int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};
In void build(int x, int y)
{
  addEdge(0, id[x][y][1], INF);
  addEdge(id[x][y][r + 1], t, INF);
  for(int k = 1; k <= r; ++k)
    {
      addEdge(id[x][y][k], id[x][y][k + 1], a[x][y][k]);
      for(int i = 0 ; i < 4; ++i)
	{
	  int nx = x + dx[i], ny = y + dy[i];
	  if(!nx || nx > p || !ny || ny > q) continue;
	  addEdge(id[x][y][k], id[nx][ny][max(1, k - d)], INF);
	  addEdge(id[nx][ny][min(r + 1, k + d + 1)], id[x][y][k + 1], INF);
	}
    }
}

int main()
{
  //freopen("ac.out", "w", stdout);
  Mem(head, -1);
  p = read(), q = read(), r = read(), d = read();
  for(int k = 1; k <= r; ++k)
    for(int i = 1; i <= p; ++i)
      for(int j = 1; j <= q; ++j) a[i][j][k] = read();
  for(int i = 1; i <= p; ++i)
    for(int j = 1; j <= q; ++j)
      for(int k = 1; k <= r + 1; ++k) id[i][j][k] = ++cnt;
  t = ++cnt;
  for(int i = 1; i <= p; ++i)
    for(int j = 1; j <= q; ++j) build(i, j);
  write(minCut()), enter;
  return 0;
}