【BZOJ3996】[TJOI2015]线性代数（最小割）

洛谷

题意：
给出一个\(n*n\)的矩阵\(B\)，再给出一个\(1*n\)的矩阵\(C\)。
求一个\(1*n\)的\(01\)矩阵\(A\)，使得\(D=(A\cdot B-C)\cdot A^T\)最大。

思路：
化简最后得：
\[ \sum_{i=1}^n\sum_{j=1}^nB_{i,j}A_iA_j-\sum_{i=1}^nA_iC_i \]
之后考虑所有的\(A_i\)都为\(1\)，现在要将一部分\(A_i\)变为\(0\)，最后的损失最小。
因为最后的\(A\)为\(01\)矩阵，显然最后结果为一个集合；结合损失最小，考虑最小割。
式子中只与\(A_i,A_j\)两个元素有关，单独考虑这两个元素，发现是一个很经典的最小割模型。然后解下方程就没了。

代码如下：

/*
 * Author:  heyuhhh
 * Created Time:  2019/10/29 17:17:13
 */
#include <bits/stdc++.h>
#define MP make_pair
#define fi first
#define se second
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#ifdef Local
  #define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
  void err() { std::cout << '\n'; }
  template<typename T, typename...Args>
  void err(T a, Args...args) { std::cout << a << ' '; err(args...); }
#else
  #define dbg(...)
#endif
void pt() {std::cout << '\n'; }
template<typename T, typename...Args>
void pt(T a, Args...args) {std::cout << a << ' '; pt(args...); }
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 500 + 5;

int n;
int a[N][N];
template <class T>
struct Dinic{
    struct Edge{
        int v, next;
        T flow;
        Edge(){}
        Edge(int v, int next, T flow) : v(v), next(next), flow(flow) {}
    }e[N * N * 10];
    int head[N], tot;
    int dep[N];
    void init() {
        memset(head, -1, sizeof(head)); tot = 0;
    }
    void adde(int u, int v, T w, T rw = 0) {
        e[tot] = Edge(v, head[u], w);
        head[u] = tot++;
        e[tot] = Edge(u, head[v], rw);
        head[v] = tot++;
    }
    bool BFS(int _S, int _T) {
        memset(dep, 0, sizeof(dep));
        queue <int> q; q.push(_S); dep[_S] = 1;
        while(!q.empty()) {
            int u = q.front(); q.pop();
            for(int i = head[u]; ~i; i = e[i].next) {
                int v = e[i].v;
                if(!dep[v] && e[i].flow > 0) {
                    dep[v] = dep[u] + 1;
                    q.push(v);
                }
            }
        }
        return dep[_T] != 0;
    }
    T dfs(int _S, int _T, T a) {
        T flow = 0, f;
        if(_S == _T || a == 0) return a;
        for(int i = head[_S]; ~i; i = e[i].next) {
            int v = e[i].v;
            if(dep[v] != dep[_S] + 1) continue;
            f = dfs(v, _T, min(a, e[i].flow));
            if(f) {
                e[i].flow -= f;
                e[i ^ 1].flow += f;
                flow += f;
                a -= f;
                if(a == 0) break;
            }
        }
        if(!flow) dep[_S] = -1;
        return flow;
    }
    T dinic(int _S, int _T) {
        T max_flow = 0;
        while(BFS(_S, _T)) max_flow += dfs(_S, _T, INF);
        return max_flow;
    }
};

Dinic <int> solver;

int c[N], v[N];

void run() {
    solver.init();
    cin >> n;
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= n; j++) {
            cin >> a[i][j];
        }
    }
    for(int i = 1; i <= n; i++) cin >> c[i];
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= n; j++) v[i] += a[i][j] + a[j][i];
        v[i] -= a[i][i];
        v[i] <<= 1;
    }
    int S = 0, T = n + 1;
    for(int i = 1; i <= n; i++) {
        int t = 0;
        for(int j = 1; j <= n; j++) {
            if(i != j) { 
                solver.adde(i, j, a[i][j] + a[j][i]);
                t += v[i] - a[i][j] - a[j][i];
            }
        }
        solver.adde(i, T, 2 * c[i]);
        solver.adde(S, i, t);
    }
    int ans = 0;
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= n; j++) {
            ans += a[i][j];
        }
    }
    int flow = solver.dinic(S, T);
    dbg(ans, flow);
    cout << (ans - flow / 2) << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    cout << fixed << setprecision(20);
    freopen("../input.in", "r", stdin);
    freopen("../output.out", "w", stdout);
    run();
    return 0;
}