P4313 最小割-CFANZ编程社区

题意

传送门 P4313 文理分科

题解

记图的 $s - t$ 割中，包含源点 $s$ 的集合为 $S$ ，包含汇点 $t$ 的集合为 $T$ 。假设被 $S$ 包含的节点选择文科，反之选择理科。将问题从求解最大值转换为求解最小值

$\sum\limits_{i,j} (art_{i,j} + science_{i,j} + same\_art_{i,j} + same\_science_{i,j}) \\- (\sum\limits_{(i,j)\in T}art_{i,j}+ \sum\limits_{(i,j)\in S}science_{i,j}+ \sum\limits_{(i,j)与相邻格子至少存在一个属于T}same\_art_{i,j}+\sum\limits_{(i,j)与相邻格子至少存在一个属于S}same\_science_{i,j})$ 最小化后一项即可。

问题转化为如何建边，使花费的总和等于割的总容量。令 $(u, v, c)$ 代表 $u\rightarrow v$ 连一条容量为 $c$ 的边。令 $(i, j)$ 代表的节点编号为 $i\times m +j$ 。对于
$\sum\limits_{(i,j)\in T}art_{i,j}$ 建边 $(s,i\times m +j,art_{i,j})$ 。对于
$\sum\limits_{(i,j)\in S}science_{i,j}$ 建边 $(i\times m +j,t,science_{i,j})$ 。对于
$\sum\limits_{(i,j)与相邻格子至少存在一个属于T}same\_art_{i,j}$ 设 $a d j$ 代表 $(i, j)$ 与相邻格子构成的点集，新建虚节点 $v$ ，建边 $s,v,same\_art_{i,j})$ 与 $(v,u,\infty),u\in adj$ 。对于最后一项同理。

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
const int MAXN = 105, MAXV = 3E4 + 5, INF = 0x3f3f3f3f;
int N, M, V;
int A[MAXN][MAXN], B[MAXN][MAXN];
int _A[MAXN][MAXN], _B[MAXN][MAXN];
struct edge
{
    int to, cap, rev;
};
vector<edge> G[MAXV];
int iter[MAXV], level[MAXV];

void add_edge(int from, int to, int cap)
{
    G[from].pb({to, cap, G[to].size()});
    G[to].pb({from, 0, G[from].size() - 1});
}

void bfs(int s)
{
    for (int v = 0; v < V; ++v)
        level[v] = -1;
    queue<int> q;
    level[s] = 0;
    q.push(s);
    while (q.size())
    {
        int v = q.front();
        q.pop();
        for (auto &e : G[v])
            if (e.cap > 0 && level[e.to] == -1)
                level[e.to] = level[v] + 1, q.push(e.to);
    }
}

int dfs(int v, int t, int f)
{
    if (v == t)
        return f;
    for (int &i = iter[v]; i < G[v].size(); ++i)
    {
        auto &e = G[v][i];
        if (e.cap > 0 && level[e.to] > level[v])
        {
            int d = dfs(e.to, t, min(f, e.cap));
            if (d > 0)
            {
                e.cap -= d;
                G[e.to][e.rev].cap += d;
                return d;
            }
        }
    }
    return 0;
}

int max_flow(int s, int t)
{
    int flow = 0;
    for (;;)
    {
        bfs(s);
        if (level[t] == -1)
            return flow;
        for (int v = 0; v < V; ++v)
            iter[v] = 0;
        int f;
        while ((f = dfs(s, t, INF)) > 0)
            flow += f;
    }
}

const int dx[5] = {0, 0, -1, 1, 0}, dy[5] = {-1, 1, 0, 0, 0};

int main()
{
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> N >> M;
    for (int i = 0; i < N; ++i)
        for (int j = 0; j < M; ++j)
            cin >> A[i][j];
    for (int i = 0; i < N; ++i)
        for (int j = 0; j < M; ++j)
            cin >> B[i][j];
    for (int i = 0; i < N; ++i)
        for (int j = 0; j < M; ++j)
            cin >> _A[i][j];
    for (int i = 0; i < N; ++i)
        for (int j = 0; j < M; ++j)
            cin >> _B[i][j];
    int res = 0;
    for (int i = 0; i < N; ++i)
        for (int j = 0; j < M; ++j)
            res += A[i][j] + B[i][j] + _A[i][j] + _B[i][j];
    int s = N * M, t = s + 1;
    V = t + 1;
    for (int i = 0; i < N; ++i)
        for (int j = 0; j < M; ++j)
            add_edge(s, i * M + j, A[i][j]), add_edge(i * M + j, t, B[i][j]);
    for (int x = 0; x < N; ++x)
        for (int y = 0; y < M; ++y)
        {
            int v = V++, u = V++;
            add_edge(s, v, _A[x][y]);
            add_edge(u, t, _B[x][y]);
            for (int d = 0; d < 5; ++d)
            {
                int nx = x + dx[d], ny = y + dy[d];
                if (0 <= nx && nx < N && 0 <= ny && ny < M)
                {
                    int w = nx * M + ny;
                    add_edge(v, w, INF);
                    add_edge(w, u, INF);
                }
            }
        }
    res -= max_flow(s, t);
    cout << res << '\n';
    return 0;
}