BZOJ3996 [TJOI2015]线性代数

就是求$D = A imes B imes A^T - C imes A^T$

展开也就是$$D = sum_{i, j} A_i * A_j * B_{i, j} - sum_{i} C_i * A_i$$

其中$Ai = 0 or 1$

转化成最小割模型，就是一堆东西，选了$i$号物品要支付费用$C_i$，同时选$i$和$j$两个物品可以获得$B_{i, j}$的收益

于是非常简单啦，建图什么的直接看程序好了！

/**************************************************************
    Problem: 3996
    User: rausen
    Language: C++
    Result: Accepted
    Time:476 ms
    Memory:51612 kb
****************************************************************/
 
#include <cstdio>
#include <cstring>
#include <algorithm>
 
using namespace std;
const int N = 505 * 505;
const int M = N << 4;
const int inf = 1e9;
 
struct edge {
    int next, to, f;
    edge() {}
    edge(int _n, int _t, int _f) : next(_n), to(_t), f(_f) {}
} e[M];
 
int n, ans, S, T;
int first[N], tot = 1;
int d[N], q[N];
 
inline int read();
 
inline void Add_Edges(int x, int y, int f) {
    e[++tot] = edge(first[x], y, f), first[x] = tot;
    e[++tot] = edge(first[y], x, 0), first[y] = tot;
}
 
#define y e[x].to
#define p q[l]
bool bfs() {
    int l, r, x;
    memset(d, -1, sizeof(d));
    d[q[1] = S] = 1;
    for (l = r = 1; l != r + 1; ++l)
        for (x = first[p]; x; x = e[x].next)
            if (!~d[y] && e[x].f) {
                d[q[++r] = y] = d[p] + 1;
                if (y == T) return 1;
            }
    return 0;
}
#undef p
 
int dfs(int p, int lim) {
  if (p == T || !lim) return lim;
  int x, tmp, rest = lim;
  for (x = first[p]; x && rest; x = e[x].next) 
    if (d[y] == d[p] + 1 && ((tmp = min(e[x].f, rest)) > 0)) {
      rest -= (tmp = dfs(y, tmp));
      e[x].f -= tmp, e[x ^ 1].f += tmp;
      if (!rest) return lim;
    }
  if (rest) d[p] = -1;
  return lim - rest;
}
#undef y
 
int Dinic() {
  int res = 0;
  while (bfs())
    res += dfs(S, inf);
  return res;
}
 
int main() {
    int i, j, x, now;
    n = now = read(), S = n * n + n + 1, T = S + 1;
    for (i = 1; i <= n; ++i)
        for (j = 1; j <= n; ++j) {
            x = read(), ++now;
            Add_Edges(i, now, inf), Add_Edges(j, now, inf);
            Add_Edges(now, T, x);
            ans += x;
        }
    for (i = 1; i <= n; ++i) Add_Edges(S, i, read());
    printf("%d
", ans - Dinic());
    return 0;
}
 
inline int read() {
    static int x;
    static char ch;
    x = 0, ch = getchar();
    while (ch < '0' || '9' < ch)
        ch = getchar();
    while ('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x;
}