zoj 2071 Technology Trader 最大权闭合子图

传送门

和上一题一样， 也是一个最大权闭合子图。不过建图好麻烦的感觉  写了好久。

源点和原材料连边， 权值为val。 汇点和产品连边， 权值为val。 产品与和它有关系的材料连边， 权值inf。 最后跑一边网络流， 满流的话是没利润的， 输出结果是利润和-最大流。

#include<bits/stdc++.h>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, a, n) for(int i = a; i<n; i++)
#define ull unsigned long long
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
const int maxn = 20000;
int head[maxn*4], s, t, num, q[maxn*10], dis[maxn], used[maxn], cnt1, cnt2, ans1[maxn], ans2[maxn];
map <string, int> vis;
struct
{
    int val;
    string name;
}a[maxn];
struct node
{
    int to, nextt, c;
}e[maxn*4];
void init() {
    mem1(head);
    num = cnt1 = cnt2 = 0;
    mem(used);
    vis.clear();
}
void add(int u, int v, int c) {
    e[num].to = v; e[num].nextt = head[u]; e[num].c = c; head[u] = num++;
    e[num].to = u; e[num].nextt = head[v]; e[num].c = 0; head[v] = num++;
}
int bfs() {
    int u, v, st = 0, ed = 0;
    mem(dis);
    dis[s] = 1;
    q[ed++] = s;
    while(st<ed) {
        u = q[st++];
        for(int i = head[u]; ~i; i = e[i].nextt) {
            v = e[i].to;
            if(e[i].c&&!dis[v]) {
                dis[v] = dis[u]+1;
                if(v == t)
                    return 1;
                q[ed++] = v;
            }
        }
    }
    return 0;
}
int dfs(int u, int limit) {
    if(u == t)
        return limit;
    int cost = 0;
    for(int i = head[u]; ~i; i = e[i].nextt) {
        int v = e[i].to;
        if(e[i].c&&dis[u] == dis[v]-1) {
            int tmp = dfs(v, min(limit-cost, e[i].c));
            if(tmp>0) {
                e[i].c -= tmp;
                e[i^1].c += tmp;
                cost += tmp;
                if(cost == limit)
                    break;
            } else {
                dis[v] = -1;
            }
        }
    }
    return cost;
}
int dinic() {
    int ans = 0;
    while(bfs()) {
        ans += dfs(s, inf);
    }
    return ans;
}
void dfs(int u) {
    used[u] = 1;
    for(int i = head[u]; ~i; i = e[i].nextt) {
        int v = e[i].to;
        if(e[i].c&&!used[v]) {
            dfs(v);
        }
    }
}
int main()
{
    int T, n, q, val, x;
    cin>>T;
    while(T--) {
        cin>>n;
        init();
        s = 0;
        int i, sum = 0;
        for(i = 1; i<=n; i++) {
            cin>>a[i].name>>a[i].val;
            vis[a[i].name] = i;
            add(s, i, a[i].val);
        }
        cin>>q;
        t = n+q+1;
        string str;
        while(q--) {
            cin>>a[i].name>>a[i].val>>x;
            sum += a[i].val;
            add(i, t, a[i].val);
            while(x--) {
                cin>>str;
                add(vis[str], i, inf);
            }
            i++;
        }
        int ans =  sum - dinic();
        dfs(0);
        for(int i = head[s]; ~i; i = e[i].nextt) {
            int v = e[i].to;
            if(!used[v]) {
                ans1[cnt1++] = v;
            }
        }
        for(int i = head[t]; ~i; i = e[i].nextt) {
            int v = e[i].to;
            if(!used[v]) {
                ans2[cnt2++] = v;
            }
        }
        cout<<ans<<endl;
        cout<<cnt2<<endl;
        for(int i = 0; i<cnt2; i++) {
            cout<<a[ans2[i]].name<<endl;
        }
        cout<<cnt1<<endl;
        for(int i = 0; i<cnt1; i++)
            cout<<a[ans1[i]].name<<endl;
    }
}