# BZOJ 2039: [2009国家集训队]employ人员雇佣

/*
BZOJ 2039: [2009国家集训队]employ人员雇佣

最小割
先全部雇佣
每个人向汇点连边

后源点分别向i,j连边
i,j之间连容量为2 * a[i][j]的边

如果用两个，就不用割
用一个或都不用，就要割掉2 *a[i][j]
答案即为利润之和减去割掉的值

*/
#include <cstdio>
#include <iostream>
#define rg register

inline void read (int &n) {
rg char c = getchar ();
for (n = 0; !isdigit (c); c = getchar ());
for (; isdigit (c); n = n * 10 + c - '0', c = getchar ());
}
#define Max 1020
int S, T;
#define INF 1e9
namespace net {

const int MaxE = 4000100;
int _n[MaxE], _v[MaxE], list[Max], _f[MaxE], EC = 1, d[Max], q[Max], tc[Max];

inline void In (int u, int v, int f) {
_v[++ EC] = v, _n[EC] = list[u], list[u] = EC, _f[EC] = f;
_v[++ EC] = u, _n[EC] = list[v], list[v] = EC, _f[EC] = 0;
}

bool Bfs () {
int h = 1, t = 1; q[t] = S; rg int i, n;
for (i = 0; i <= T; ++ i) d[i] = -1;
for (d[S] = 0; h <= t; ++ h)
for (n = q[h], i = list[n]; i; i = _n[i])
if (_f[i] && d[_v[i]] < 0) {
d[_v[i]] = d[n] + 1, q[++ t] = _v[i];
if (_v[i] == T) return true;
}
return false;
}
int Flowing (int n, int f) {
if (n == T || f == 0) return f;
int p, r = 0;
for (rg int &i = tc[n]; i; i = _n[i])
if (_f[i] && d[_v[i]] == d[n] + 1) {
p = Flowing (_v[i], std :: min (_f[i], f));
if (p > 0) {
r += p, f -= p, _f[i] -= p, _f[i ^ 1] += p;
if (f == 0) return r;
}
}
if (r != f) d[n] = -1; return r;
}

int Dinic () {
int res = 0;
for (; Bfs (); res += Flowing (S, INF))
for (rg int i = 0; i <= T; ++ i) tc[i] = list[i];
return res;
}
}
int a[Max][Max];
int main (int argc, char *argv[]) {
int N, x, res = 0; read (N); S = 0, T = N + 1; rg int i, j;
for (i = 1; i <= N; ++ i) read (x), net :: In (i, T, x);
for (i = 1; i <= N; ++ i)
for (j = 1; j <= N; ++ j) read (a[i][j]), res += a[i][j];

for (i = 1; i <= N; ++ i)
for (j = 1 + i; j <= N; ++ j)
net :: In (S, i, a[i][j]), net :: In (S, j, a[i][j]), net :: In (i, j, a[i][j]<< 1), net :: In (j, i, a[i][j] << 1);

printf ("%d", res - net :: Dinic ());
return 0;
}

posted @ 2018-01-19 21:07  ZlycerQan  阅读(...)  评论(...编辑  收藏