「雅礼集训 2017 Day8」价

#6045. 「雅礼集训 2017 Day8」价

Summarize

有 $n$ 种药,每种药由若干药材组成,恰好有 $n$ 种不同的药材。要求选出 $k$ 种药,并且使用的药材并集大小也为 $k$,使得药材的权值和最小。

$n \le 300$

Solution

二分图建模,不同的药向对应的药材连边,边权为无穷大;源点向药连边,边权为 $-p_i$;药材向汇点连边,边权为 $0$。在原图上求最小闭合子图即可。由于存在负权,可以将边权加上 $1e9$ 后再进行操作。

Code

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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const ll INF = 1e9;
const int MAXN = 605;
const int MAXM = 605 * 305;

int n, m, s, t;

struct Edge {
int v, nxt;
ll w;
} edge[MAXM << 1];
int head[MAXN], cnt = 1;
int head_or[MAXN];

void AddEdge(int u, int v, ll w) {
edge[++cnt].w = w;
edge[cnt].v = v;
edge[cnt].nxt = head[u];
head[u] = cnt;
}

int d[MAXN];
bool vis[MAXN];

bool bfs() {
memset(vis, 0, sizeof(vis));
memcpy(head, head_or, sizeof(head));
vis[s] = 1, d[s] = 0;
queue<int> q;
q.push(s);
while (q.size()) {
int u = q.front();
q.pop();
for (int i = head[u]; i; i = edge[i].nxt) {
int v = edge[i].v;
ll w = edge[i].w;
if (vis[v] || (w == 0)) continue;
vis[v] = 1;
d[v] = d[u] + 1;
q.push(v);
}
}
return vis[t];
}

ll dfs(int u, ll flow) {
if (u == t) return flow;
ll rest = flow;
for (int i = head[u]; i && rest; i = edge[i].nxt) {
head[u] = i;
int v = edge[i].v;
ll w = edge[i].w;
if (w == 0 || d[v] != d[u] + 1) continue;
ll tmp = dfs(v, min(rest, w));
if (tmp == 0) d[v] = 0;
edge[i].w -= tmp;
edge[i ^ 1].w += tmp;
rest -= tmp;
}
return flow - rest;
}

ll dinic() {
ll ret = 0, tmp = 0;
while (bfs())
while (tmp = dfs(s, INF)) ret += tmp;
return ret;
}

int main() {
scanf("%d", &n);
s = 2 * n + 1, t = 2 * n + 2;
for (int i = 1; i <= n; ++i) {
int k = 0;
scanf("%d", &k);
for (int j = 1; j <= k; ++j) {
int x;
scanf("%d", &x);
AddEdge(i, n + x, INF);
AddEdge(n + x, i, 0);
}
}
ll ans2 = 0;
for (int i = 1; i <= n; ++i) {
int p;
scanf("%d", &p);
ans2 += INF - p;
AddEdge(s, i, INF - p);
AddEdge(i, s, 0);
AddEdge(n + i, t, INF);
AddEdge(t, n + i, 0);
}
memcpy(head_or, head, sizeof(head_or));
ll ans = dinic();
cout << ans - ans2 << endl;

return 0;
}

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