BZOJ3996 [TJOI2015]线性代数

Description

给定\(n*n\)的矩阵\(B\)和\(1*n\)的矩阵\(C\)。求一个\(1*N\)的01矩阵\(A\)使得\((AB-C)A^T\)最大。\(A^T\)是\(A\)的转置。\(n\leq500\)，所有输入为不超过\(1000\)的非负整数。

Solution

\[\begin{aligned} (AB)_i&=\sum_{j=1}^n A_jB_{j,i}\\ (AB-C)_i&=\left(\sum_{j=1}^n A_jB_{j,i}\right) - C_i\\ (AB-C)A^T&=\sum_{i=1}^n A_i\left(\sum_{j=1}^nA_jB_{j,i}\right)-A_iC_i\\ &=\sum_{i,j=1\dots n} A_iA_jB_{i,j} -\sum_{i=1}^nA_iC_i \end{aligned} \]

观察最后一个式子，可以看成：

\(n\)个物品，第\(i,j\)两个物品同时选会获得\(B_{i,j}\)的收益；选第\(i\)个物品会付出\(C_i\)的代价；求最大净收益。

最小割即可。

Code

#include <algorithm>
#include <cstdio>
#include <cstring>
const int N = 500;
const int NN = 500000;
const int M = 3000050;
int pre[NN], nxt[M], to[M], ret[M], cnt;
int dis[NN], que[NN];
bool BFS(int S, int T) {
  int hd = 0, tl = 0;
  memset(dis, -1, sizeof dis);
  for (dis[que[tl++] = S] = 0; hd < tl; ++hd)
    for (int i = pre[que[hd]]; ~i; i = nxt[i])
      if (ret[i] && !~dis[to[i]]) dis[que[tl++] = to[i]] = dis[que[hd]] + 1;
  return dis[T] != -1;
}
int DFS(int x, int T, int maxf) {
  if (!maxf) return 0;
  if (x == T) return maxf;
  int ans = 0;
  for (int i = pre[x]; ~i; i = nxt[i])
    if (ret[i] && dis[to[i]] == dis[x] + 1) {
      int t = DFS(to[i], T, std::min(maxf - ans, ret[i]));
      ret[i] -= t; ret[i ^ 1] += t;
      ans += t;
    }
  if (ans < maxf) dis[x] = -1;
  return ans;
}
int solve(int S, int T) {
  int ans = 0;
  while (BFS(S, T)) ans += DFS(S, T, 1000000000);
  return ans;
}
inline void addEdge(int x, int y, int c) {
  nxt[cnt] = pre[x];
  ret[cnt] = c;
  to[pre[x] = cnt++] = y;
  nxt[cnt] = pre[y];
  ret[cnt] = 0;
  to[pre[y] = cnt++] = x;
}
int main() {
  int n;
  scanf("%d", &n);
  memset(pre, -1, sizeof pre);
  int S = n * (n + 1), T = S + 1;
  int ans = 0;
  for (int i = 0; i < n; ++i)
    for (int j = 0, v; j < n; ++j) {
      scanf("%d", &v);
      ans += v;
      int t = (i + 1) * n + j;
      addEdge(t, T, v);
      addEdge(i, t, 1000000000);
      addEdge(j, t, 1000000000);
    }
  for (int i = 0, v; i < n; ++i) {
    scanf("%d", &v);
    addEdge(S, i, v);
  }
  printf("%d\n", ans - solve(S, T));
  return 0;
}