[最小割] BZOJ 3144: [Hnoi2013]切糕

来源：互联网发布：sql去除查询结果重复行编辑：程序博客网时间：2024/06/03 08:55

Solution

建立超级源汇，超级源连到切糕的顶部，切糕的底部连到超级汇，流量无穷大。
对于(x,y,∗)从上到下连边，这里的边就相当于原来要切的点，标上流量。
如果不考虑f(x,y)−f(x′,y′)≤D的话，最小割就是答案。
那么可以这么连。
这里写图片描述
如果选取的点不在区间[z−D,z+D]里就会存在无穷大的边无法割掉。

#include <bits/stdc++.h>using namespace std;const int N = 50;const int INF = 1 << 29;inline char get(void) {  static char buf[100000], *S = buf, *T = buf;  if (S == T) {    T = (S = buf) + fread(buf, 1, 100000, stdin);    if (S == T) return EOF;  }  return *S++;}template<typename T>inline void read(T &x) {  static char c; x = 0; int sgn = 0;  for (c = get(); c < '0' || c > '9'; c = get()) if (c == '-') sgn = 0;  for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';  if (sgn) x = -x;}struct edge {  int to, next, flow;  edge(int t = 0, int n = 0, int f = 0):to(t), next(n), flow(f) {}};edge G[N * N * N * 10];int head[N * N * N], cur[N * N * N];int Gcnt, p, q, r, d, S, T, cnt, clc;int v[N][N][N], id[N][N][N];int dis[N * N * N], vis[N * N * N];queue<int> Q;inline void AddEdge(int from, int to, int flow) {  G[++Gcnt] = edge(to, head[from], flow); head[from] = Gcnt;  G[++Gcnt] = edge(from, head[to], 0); head[to] = Gcnt;  // printf("%d %d %d\n", from, to, flow);}inline bool bfs(int S, int T) {  vis[S] = ++clc; dis[S] = 0;  Q.push(S); dis[T] = INF;  while (!Q.empty()) {    int u = Q.front(); Q.pop();    for (int i = head[u]; i; i = G[i].next) {      edge &e = G[i];      if (e.flow && vis[e.to] != clc) {    dis[e.to] = dis[u] + 1;    Q.push(e.to); vis[e.to] = clc;      }    }  }  return dis[T] != INF;}inline int dfs(int u, int a) {  if (u == T || !a) return a;  int f, flow = 0;  for (int &i = cur[u]; i; i = G[i].next) {    edge &e = G[i];    if (dis[e.to] == dis[u] + 1 && e.flow    && (f = dfs(e.to, min(e.flow, a))) > 0) {      e.flow -= f; G[i ^ 1].flow += f;      flow += f; a -= f; if (!a) break;    }  }  return flow;}inline int MaxFlow(int S, int T) {  int flow = 0;  while (bfs(S, T)) {    for (int i = 1; i <= T; i++) cur[i] = head[i];    flow += dfs(S, INF);  }  return flow;}int main(void) {  Gcnt = 1;  read(p); read(q); read(r); read(d);  for (int i = 1; i <= r; i++)    for (int j = 1; j <= p; j++)      for (int k = 1; k <= q; k++)    read(v[i][j][k]);  for (int i = 0; i <= r; i++)    for (int j = 1; j <= p; j++)      for (int k = 1; k <= q; k++)    id[i][j][k] = ++cnt;  S = ++cnt; T = ++cnt;  for (int j = 1; j <= p; j++)    for (int k = 1; k <= q; k++) {      AddEdge(S, id[0][j][k], INF);      AddEdge(id[r][j][k], T, INF);      for (int i = 1; i <= r; i++) {    AddEdge(id[i - 1][j][k], id[i][j][k], v[i][j][k]);    for (int x = -1; x <= 1; x++)      for (int y = -1; y <= 1; y++) {        if (!(x && !y) && !(y && !x)) continue;        if (j + x < 1 || j + x > p) continue;        if (k + y < 1 || k + y > q) continue;        if (i > d) AddEdge(id[i][j][k], id[i - d][j + x][k + y], INF);      }      }    }  printf("%d\n", MaxFlow(S, T));  return 0;}

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