BZOJ3996 [TJOI2015]线性代数

就是求$D = A \times B \times A^T - C \times A^T$

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 =  * ;
 const int M = N << ;
 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 = ;
 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, ), first[y] = tot;
 }

 #define y e[x].to
 #define p q[l]
 bool bfs() {
     int l, r, x;
     memset(d, -, sizeof(d));
     d[q[] = S] = ;
     for (l = r = ; l != r + ; ++l)
         for (x = first[p]; x; x = e[x].next)
             if (!~d[y] && e[x].f) {
                 d[q[++r] = y] = d[p] + ;
                 if (y == T) return ;
             }
     return ;
 }
 #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] +  && ((tmp = min(e[x].f, rest)) > )) {
       rest -= (tmp = dfs(y, tmp));
       e[x].f -= tmp, e[x ^ ].f += tmp;
       if (!rest) return lim;
     }
   if (rest) d[p] = -;
   return lim - rest;
 }
 #undef y

 int Dinic() {
   int res = ;
   while (bfs())
     res += dfs(S, inf);
   return res;
 }

 int main() {
     int i, j, x, now;
     n = now = read(), S = n * n + n + , T = S + ;
     for (i = ; i <= n; ++i)
         for (j = ; 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 = ; i <= n; ++i) Add_Edges(S, i, read());
     printf("%d\n", ans - Dinic());
     return ;
 }

 inline int read() {
     static int x;
     static char ch;
     x = , ch = getchar();
     while (ch < '' || '' < ch)
         ch = getchar();
     while ('' <= ch && ch <= '') {
         x = x *  + ch - '';
         ch = getchar();
     }
     return x;
 }