【网络流 24 题】方格取数（二分图的最大点权独立集）

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题意

在一个有 $m \times n$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。

题解

题目要求不相邻，可以转换为最大独立集，又由于点权不全为1，则为最大点权独立集。
最大点权独立集 = 总点权 - 最小点权覆盖，问题转换为如何求最小点权覆盖。
建立二分图，源点向左点集连边， 容量为点权。右点集向汇点连边，容量为点权。内部的点连边，容量为INF。跑最大流，求出最小割即可。

代码

#include<bits/stdc++.h>
using namespace std;
typedef double db;
typedef long long ll;
typedef unsigned long long ull;
const int nmax = 1e6+7;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const ull p = 67;
const ull MOD = 1610612741;
int n, m;
int mp[50][50];
struct Dinic {
    int head[nmax], cur[nmax], d[nmax];
    bool vis[nmax];
    int tot, n, m, s, t, front, tail;
    int qqq[nmax];
    struct edge {
        int nxt, to, w, cap, flow;
    } e[nmax<<1];
    void init(int n) {
        this->n = n;
        memset(head, -1, sizeof head);
        memset(cur, 0, sizeof cur);
        memset(e,0,sizeof e);
        this->tot = 0;
    }
    int add_edge(int u, int v, int c) {
        int temp = tot;
        e[tot].to = v, e[tot].cap = c, e[tot].flow = 0;
        e[tot].nxt = head[u];
        head[u] = tot++;
        e[tot].to = u, e[tot].cap = c, e[tot].flow = c;
        e[tot].nxt = head[v];
        head[v] = tot++;
//        printf("add %d %d %d\n", u, v, c);
        return temp;
    }
    bool BFS() {
        for(int i = 0; i <= n; ++i) vis[i] = false;
        front = tail = 0;
        vis[s] = 1; d[s] = 0;
        qqq[tail++] = s;
        while (front < tail) {
            int u = qqq[front++];
            for (int i = head[u]; i != -1; i = e[i].nxt) {
                int v = e[i].to;
                if (!vis[v] && e[i].cap > e[i].flow) {
                    vis[v] = 1;
                    d[v] = d[u] + 1;
                    qqq[tail++] = v;
                }
            }
        }
        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 != -1; i = e[i].nxt) {
            int v = e[i].to;
            if (d[v] == d[x] + 1 && (f = DFS(v, min(a, e[i].cap - e[i].flow))) > 0) {
                Flow += f;
                e[i].flow += f;
                e[i ^ 1].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()) {
            for (int i = 0; i <= n; i++) cur[i] = head[i];
            Flow += DFS(s,INF);
        }
        return Flow;
    }
} dinic;
int main(){
    scanf("%d %d", &n, &m);
    int sum = 0;
    int s = 0, t = n * m + 1;
    dinic.init(t);
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            scanf("%d", &mp[i][j]);
            sum += mp[i][j];

        }
    }
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            if ((i + j) % 2 == 1) {
                dinic.add_edge(s, (i - 1) * m + j, mp[i][j]);
                if (i != n) {
                    dinic.add_edge((i - 1) * m + j, i * m + j, INF);
                }
                if (j != m) {
                    dinic.add_edge((i - 1) * m + j, (i - 1) * m + j + 1, INF);
                }
                if (i != 1) {
                    dinic.add_edge((i - 1) * m + j, (i - 2) * m + j, INF);
                }
                if (j != 1) {
                    dinic.add_edge((i - 1) * m + j, (i - 1) * m + j - 1, INF);
                }
            } else {
                dinic.add_edge((i - 1) * m + j, t, mp[i][j]);
            }
        }
    }
    int mxflow = dinic.Maxflow(s, t);
//    printf("%d\n", mxflow);
    printf("%d\n", sum - mxflow);
    return 0;
}