1744: 方格取数问题——最小割

题意：

在一个有m*n 个方格的棋盘中，每个方格中有一个正整数。现要从方格中取数，使任 意2 个数所在方格没有公共边，且取出的数的总和最大。试设计一个满足要求的取数算法。 编程任务： 对于给定的方格棋盘，按照取数要求编程找出总和最大的数。

思路：

首先对矩阵进行黑白点染色， 相邻的两个点颜色不同。然后每个黑点向四周连有向边，容量为INF， 之后源点向黑点连有向边，容量为点权，白点向汇点连边，容量为点权。然后跑最小割，用所有点权之和减最小割即可。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>

using namespace std;

const int maxn = 1000;
const int INF = 0x3f3f3f3f;
const int dx[] = {0, 0, -1, 1};
const int dy[] = {-1, 1, 0, 0};

struct Edge {
    int from, to, cap, flow;
};

struct Dinic {
    int s, t;
    vector<Edge> edges;
    vector<int> G[maxn];
    bool vis[maxn];
    int d[maxn];
    int cur[maxn];

    void init() {
        edges.clear();
        for (int i = 0; i < maxn; i++) G[i].clear();
    }

    void addedge(int from, int to, int cap) {
        edges.push_back(Edge{from, to, cap, 0});
        edges.push_back(Edge{to, from, 0, 0});
        int x = edges.size();
        G[from].push_back(x-2);
        G[to].push_back(x-1);
    }

    bool bfs() {
        memset(vis, 0, sizeof(vis));
        queue<int> q;
        q.push(s);
        d[s] = 0;
        vis[s] = 1;
        while (!q.empty()) {
            int x = q.front(); q.pop();
            for (int i = 0; i < G[x].size(); i++) {
                Edge &e = edges[G[x][i]];
                if (!vis[e.to] && e.cap > e.flow) {
                    vis[e.to] = 1;
                    d[e.to] = d[x] + 1;
                    q.push(e.to);
                }
            }
        }
        return vis[t];
    }

    int dfs(int x, int a) {
        if (x == t || a == 0) return a;
        int flow = 0, f;
        for (int &i = cur[x]; i < G[x].size(); i++) {
            Edge &e = edges[G[x][i]];
            if (d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0) {
                e.flow += f;
                edges[G[x][i]^1].flow -= f;
                flow += f;
                a -= f;
                if (a == 0) break;
            }
        }
        return flow;
    }

    int maxflow(int s, int t) {
        this->s = s, this->t = t;
        int flow = 0;
        while (bfs()) {
            memset(cur, 0, sizeof(cur));
            flow += dfs(s, INF);
        }
        return flow;
    }
}ac;

int a[100][100], flag[100][100];

int main() {
    int m, n;
    scanf("%d%d", &m, &n);
    int s = 0, t = m*n + 1;
    int sum = 0;
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            scanf("%d", &a[i][j]);
            sum += a[i][j];
        }
    }
    memset(flag, 0, sizeof(flag));
    int f;
    for (int i = 1; i <= m; i++) {
        f = i & 1;
        for (int j = 1; j <= n; j++) {
            flag[i][j] = f;
            f ^= 1;
        }
    }
    //build
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            if (flag[i][j] == 0) continue;
            for (int k = 0; k < 4; k++) {
                int x = i + dx[k], y = j + dy[k];
                if (1 <= x && x <= m && 1 <= y && y <= n) {
                    ac.addedge(n*(i-1)+j, n*(x-1)+y, INF);
                }
            }
        }
    }
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            if (flag[i][j] == 1) ac.addedge(s, n*(i-1)+j, a[i][j]);
            else ac.addedge(n*(i-1)+j, t, a[i][j]);
        }
    }
    printf("%d\n", sum - ac.maxflow(s, t));
    return 0;
}
/*
3 3
1 2 3
3 2 3
2 3 1
*/