bzoj 1976 [BeiJing2010组队]能量魔方 Cube

题目描述

简要题意:给定一个 $n\times n\times n$ 的六连通立方体,每个位置可以填 $0$ 或 $1$,现在有些位置已经填了数,如果两个相邻的格子所填数不同则会产生 $1$ 的收益,问最大收益

$n\le 40$

Solution

注意到依然是相反选择有收益且三维六连通依然可以二分图染色

建图:

对于黑点 $(i,j,k)$

如果已经有选择,则按照选择向 $s$ 或 $t$ 连一条容量为 $\infty$ 的边

$(i,j,k)$ 连相邻白点,容量为 $1$,注意连双向边

对于黑点 $(i,j,k)$

如果已经有选择,则按照选择向 $s$ 或 $t$ 连一条容量为 $\infty$ 的边

注意要与白点相反

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#include <iostream>
#include <cstdio>
#include <queue>
#define maxn 6410
#define maxm 41
#define INF 1000000000
using namespace std;

int dx[] = { 1, -1, 0, 0, 0, 0 };
int dy[] = { 0, 0, 1, -1, 0, 0 };
int dz[] = { 0, 0, 0, 0, 1, -1 };

int n, m, s, t;

int g[maxm][maxm][maxm];

inline int D(int x, int y, int z) { return (z - 1) * n * n + (x - 1) * n + y; }

struct Edge {
int to, next, w;
} e[1000000]; int c1, head[maxn];
inline void add_edge(int u, int v, int w) {
e[c1].to = v; e[c1].w = w;
e[c1].next = head[u]; head[u] = c1++;
}

inline void Add_edge(int u, int v, int w) {
add_edge(u, v, w); add_edge(v, u, 0);
}

int dep[maxn];
bool bfs() {
fill(dep, dep + maxn, 0); dep[s] = 1;
queue<int> Q; Q.push(s);
while (!Q.empty()) {
int u = Q.front(); Q.pop();
for (int i = head[u]; ~i; i = e[i].next) {
int v = e[i].to, w = e[i].w;
if (!dep[v] && w > 0) {
dep[v] = dep[u] + 1;
Q.push(v); if (v == t) return 1;
}
}
}
return 0;
}

int dfs(int u, int _w) {
if (u == t || !_w) return _w;
int s = 0;
for (int i = head[u]; ~i; i = e[i].next) {
int v = e[i].to, w = e[i].w;
if (w > 0 && dep[v] == dep[u] + 1) {
int di = dfs(v, min(_w - s, w));
e[i].w -= di; e[i ^ 1].w += di;
s += di; if (s == _w) break;
}
}
if (s < _w) dep[u] = 0;
return s;
}

int dinic() {
int mf = 0;
while (bfs()) mf += dfs(s, INF);
return mf;
}

int main() { fill(head, head + maxn, -1);
ios::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);

cin >> n; s = 0; t = n * n * n + 1;
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= n; ++j) {
char s[maxm]; cin >> s + 1;
for (int k = 1; k <= n; ++k)
if (s[k] == 'P') g[i][j][k] = 1;
else if (s[k] == 'N') g[i][j][k] = -1;
}
int ans = 0;
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= n; ++j)
for (int k = 1; k <= n; ++k) {
int id = D(i, j, k);
if (i + j + k & 1) {
if (g[i][j][k] == 1) Add_edge(s, id, INF);
else if (g[i][j][k] == -1) Add_edge(id, t, INF);
for (int o = 0; o < 6; ++o) {
int x = i + dx[o], y = j + dy[o], z = k + dz[o];
if (x < 1 || x > n || y < 1 || y > n || z < 1 || z > n) continue;
Add_edge(id, D(x, y, z), 1);
Add_edge(D(x, y, z), id, 1);
++ans;
}
}
else {
if (g[i][j][k] == 1) Add_edge(id, t, INF);
else if (g[i][j][k] == -1) Add_edge(s, id, INF);
}
}
cout << ans - dinic() << "\n";
return 0;
}