bzoj 2039 [2009国家集训队]employ人员雇佣 最小割建图

题面

题目传送门

解法

\(ans\) = 所有的收益 - \(i\)属于\(S\)割的(雇佣\(i\)的代价) - 不能同时雇佣\(i,j\)的带价(\(i\)属于\(S\)割,\(j\)属于T割) - 不雇佣\(i\)所造成的带价也就是\(i\)所有的贡献(\(i\)属于\(T\)割)

那么建图也就出来了：\(S\rightarrow i\)连\(a_i\)，\(i\rightarrow j\)连\(E_{i,j}\)，\(i\rightarrow T\)连\(i\)的所有贡献

答案就是所有的贡献和减去最小割。

注意要加当前弧优化

代码

#include <bits/stdc++.h>
#define N 2010
using namespace std;
template <typename node> void read(node &x) {
	x = 0; int f = 1; char c = getchar();
	while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
	while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;
}
struct Edge {
	int next, num, c;
} e[N * N];
int s, t, cnt, a[N], sum[N], l[N], cur[N], v[N][N];
void add(int x, int y, int c) {
	e[++cnt] = (Edge) {e[x].next, y, c};
	e[x].next = cnt;
}
void Add(int x, int y, int c) {
	add(x, y, c), add(y, x, 0);
}
bool bfs(int s) {
	for (int i = 1; i <= t; i++) l[i] = -1;
	queue <int> q; q.push(s);
	while (!q.empty()) {
		int x = q.front(); q.pop();
		for (int p = e[x].next; p; p = e[p].next) {
			int k = e[p].num, c = e[p].c;
			if (c && l[k] == -1)
				l[k] = l[x] + 1, q.push(k);
		}
	}
	return l[t] != -1;
}
int dfs(int x, int lim) {
	if (x == t) return lim;
	int used = 0;
	for (int p = cur[x]; p; p = e[p].next) {
		int k = e[p].num, c = e[p].c;
		if (l[k] == l[x] + 1 && c) {
			int w = dfs(k, min(c, lim - used));
			e[p].c -= w, e[p ^ 1].c += w;
			if (e[p].c) cur[x] = p;
			used += w;
			if (used == lim) return used;
		}
	}
	if (!used) l[x] = -1;
	return used;
}
int dinic() {
	int ret = 0;
	while (bfs(s)) {
		for (int i = s; i <= t; i++)
			cur[i] = e[i].next;
		ret += dfs(s, INT_MAX);
	}
	return ret;
}
int main() {
	int n, ans = 0; read(n);
	for (int i = 1; i <= n; i++) read(a[i]);
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)
			read(v[i][j]), sum[i] += v[i][j], ans += v[i][j];
	s = 0, t = cnt = n + 1;
	if (cnt % 2 == 0) cnt++;
	for (int i = 1; i <= n; i++)
		Add(s, i, a[i]), Add(i, t, sum[i]);
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)
			Add(i, j, v[i][j] * 2);
	cout << ans - dinic() << "\n";
	return 0;
}