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-;
 const int mod = 1e9+;
 const int inf = ;
 const int dir[][] = { {-, }, {, }, {, -}, {, } };
 const int maxn = ;
 int head[maxn*], s, t, num, q[maxn*], 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*];
 void init() {
     mem1(head);
     num = cnt1 = cnt2 = ;
     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 = ; head[v] = num++;
 }
 int bfs() {
     int u, v, st = , ed = ;
     mem(dis);
     dis[s] = ;
     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]+;
                 if(v == t)
                     return ;
                 q[ed++] = v;
             }
         }
     }
     return ;
 }
 int dfs(int u, int limit) {
     if(u == t)
         return limit;
     int cost = ;
     for(int i = head[u]; ~i; i = e[i].nextt) {
         int v = e[i].to;
         if(e[i].c&&dis[u] == dis[v]-) {
             int tmp = dfs(v, min(limit-cost, e[i].c));
             if(tmp>) {
                 e[i].c -= tmp;
                 e[i^].c += tmp;
                 cost += tmp;
                 if(cost == limit)
                     break;
             } else {
                 dis[v] = -;
             }
         }
     }
     return cost;
 }
 int dinic() {
     int ans = ;
     while(bfs()) {
         ans += dfs(s, inf);
     }
     return ans;
 }
 void dfs(int u) {
     used[u] = ;
     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 = ;
         int i, sum = ;
         for(i = ; 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+;
         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();
         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 = ; i<cnt2; i++) {
             cout<<a[ans2[i]].name<<endl;
         }
         cout<<cnt1<<endl;
         for(int i = ; i<cnt1; i++)
             cout<<a[ans1[i]].name<<endl;
     }
 }