题意
给出一个 (N imes N) 的矩阵 (B) 和一个 (1 imes N) 的矩阵 (C)。
求出一个 (1 imes N) 的 (01) 矩阵 (A),使得 (D = (A imes B−C)×A^T)最大。
输出 (D)。
思路
先对式子进行化简:
不妨设(B = (a_{ij})_{n imes n}),(C = (c_i)_n)
[egin{aligned}
D &= A imes B imes A^T - C imes A^T \
&= sum_{i = 1}^nsum_{j = 1}^na_{ij}x_ix_j - sum_{i = 1}^nc_ix_i
end{aligned}
]
发现这道题就是最大获利那道题的变形,考虑最大密度子图
两点之间的边权为(a_{ij} + a_{ji}),点权为(-c_i + a_{ii})
代码
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int N = 510, M = (250000 + N * 2) * 2, inf = 1e8;
int n, S, T;
int h[N], e[M], ne[M], f[M], idx;
int cur[N], d[N];
int dg[N], p[N];
int w[N][N];
void add(int a, int b, int c1, int c2)
{
e[idx] = b, f[idx] = c1, ne[idx] = h[a], h[a] = idx ++;
e[idx] = a, f[idx] = c2, ne[idx] = h[b], h[b] = idx ++;
}
bool bfs()
{
memset(d, -1, sizeof(d));
queue<int> que;
que.push(S);
d[S] = 0, cur[S] = h[S];
while(que.size()) {
int t = que.front();
que.pop();
for(int i = h[t]; ~i; i = ne[i]) {
int ver = e[i];
if(d[ver] == -1 && f[i]) {
d[ver] = d[t] + 1;
cur[ver] = h[ver];
if(ver == T) return true;
que.push(ver);
}
}
}
return false;
}
int find(int u, int limit)
{
if(u == T) return limit;
int flow = 0;
for(int i = cur[u]; ~i && flow < limit; i = ne[i]) {
cur[u] = i;
int ver = e[i];
if(d[ver] == d[u] + 1 && f[i]) {
int t = find(ver, min(f[i], limit - flow));
if(!t) d[ver] = -1;
f[i] -= t, f[i ^ 1] += t, flow += t;
}
}
return flow;
}
int dinic()
{
int res = 0, flow;
while(bfs()) {
while(flow = find(S, inf)) {
res += flow;
}
}
return res;
}
int main()
{
scanf("%d", &n);
memset(h, -1, sizeof(h));
S = 0, T = n + 1;
for(int i = 1; i <= n; i ++) {
for(int j = 1; j <= n; j ++) {
int x;
scanf("%d", &x);
w[i][j] = x;
}
}
for(int i = 1; i <= n; i ++) {
for(int j = i + 1; j <= n; j ++) {
add(i, j, w[i][j] + w[j][i], w[i][j] + w[j][i]);
dg[i] += w[i][j] + w[j][i];
dg[j] += w[i][j] + w[j][i];
}
}
for(int i = 1; i <= n; i ++) {
scanf("%d", &p[i]);
p[i] = -p[i] + w[i][i];
}
int U = 0;
for(int i = 1; i <= n; i ++) U = max(U, 2 * p[i] + dg[i]);
for(int i = 1; i <= n; i ++) add(S, i, U, 0);
for(int i = 1; i <= n; i ++) add(i, T, U - 2 * p[i] - dg[i], 0);
printf("%d
", (n * U - dinic()) / 2);
return 0;
}