叫什么"最大权闭合图" -- from PoPoQQQ
首先建立源点汇点S、T表示种在a，b两片土里，流量为收益。
然后每次读到一个集合的时候，新建两个虚拟节点si、ti，表示都种在a，b里的收益
S向si连边，ti向T连边，流量为收益，然后si、ti向集合内所有点连边。
之后跑一边最大流，ans = 总收益 - 最小割
1 /************************************************************** 2 Problem: 3438 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:2664 ms 7 Memory:24280 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cstring> 12 #include <algorithm> 13 14 using namespace std; 15 const int inf = (int) 1e9; 16 const int N = 3005; 17 const int M = 2000005; 18 19 struct edges{ 20 int next, to, f; 21 edges() {} 22 edges(int _n, int _t, int _f) : next(_n), to(_t), f(_f) {} 23 } e[M]; 24 25 int n, m; 26 int cnt, ans, S, T; 27 int first[N], tot = 1; 28 int q[N], d[N]; 29 30 inline int read() { 31 int x = 0; 32 char ch = getchar(); 33 while (ch < '0' || '9' < ch) 34 ch = getchar(); 35 while ('0' <= ch && ch <= '9') { 36 x = x * 10 + ch - '0'; 37 ch = getchar(); 38 } 39 return x; 40 } 41 42 inline void Add_Edges(int x, int y, int z){ 43 e[++tot] = edges(first[x], y, z), first[x] = tot; 44 e[++tot] = edges(first[y], x, 0), first[y] = tot; 45 } 46 47 bool bfs(){ 48 memset(d, 0, sizeof(d)); 49 q[1] = S, d[S] = 1; 50 int l = 0, r = 1, x, y; 51 while (l < r){ 52 ++l; 53 for (x = first[q[l]]; x; x = e[x].next){ 54 y = e[x].to; 55 if (!d[y] && e[x].f) 56 q[++r] = y, d[y] = d[q[l]] + 1; 57 } 58 } 59 return d[T]; 60 } 61 62 int dinic(int p, int limit){ 63 if (p == T || !limit) return limit; 64 int x, y, tmp, rest = limit; 65 for (x = first[p]; x; x = e[x].next){ 66 y = e[x].to; 67 if (d[y] == d[p] + 1 && e[x].f && rest){ 68 tmp = dinic(y, min(rest, e[x].f)); 69 rest -= tmp; 70 e[x].f -= tmp, e[x ^ 1].f += tmp; 71 if (!rest) return limit; 72 } 73 } 74 if (limit == rest) d[p] = 0; 75 return limit - rest; 76 } 77 78 int Dinic(){ 79 int res = 0, x; 80 while (bfs()) 81 res += dinic(S, inf); 82 return res; 83 } 84 85 int main() { 86 int i, j, x, y, z; 87 n = cnt = read(), S = 3001, T = 3002; 88 for (i = 1; i <= n; ++i) { 89 ans += (x = read()); 90 Add_Edges(S, i, x); 91 } 92 for (i = 1; i <= n; ++i) { 93 ans += (x = read()); 94 Add_Edges(i, T, x); 95 } 96 m = read(); 97 for (i = 1; i <= m; ++i) { 98 j = read(), x = read(), y = read(); 99 ans += x + y; 100 Add_Edges(S, ++cnt, x), Add_Edges(++cnt, T, y); 101 while (j--) { 102 z = read(); 103 Add_Edges(z, cnt, inf), Add_Edges(cnt - 1, z, inf); 104 } 105 } 106 printf("%d ", ans - Dinic()); 107 return 0; 108 }