Hdoj 4971 a simple brute force problem. [maximum closing weight --> Minimum Cut]

Question: hdoj 4971 a simple brute force problem.

Question:Given n tasks and M technologies, several technologies are required to complete a task. To complete a task, you have a bonus. Learning a technology requires money,There is a parent-child relationship between technologies. A technology may need to learn other technologies first,Then, how many tasks do you choose to achieve the greatest benefit?

Analysis: The black text part of the question is a standard closed right problem. The key to this question is the relationship between technologies, which leads many people to think of DP and tree-like DP.

In fact, it is a closed-Permission question template. The official question is: if there is a mutual relationship between technologies, it needs to be scaled down. In fact, it does not need to be scaled down.

Graph creation:

# Include <cstdio> # include <algorithm> # include <vector> # include <queue> # include <cstring> const int n = 200; using namespace STD; const int INF = 0x3f3f3f; # define del (a, B) memset (a, B, sizeof (A) struct node {int from, to, Cap, flow ;}; vector <int> V [N]; vector <node> E; int vis [N], cur [N]; void add_node (int from, int to, int cap) {e. push_back (node) {from, to, Cap, 0}); E. push_back (node) {to, from, 0, 0}); int TMP = E. size (); V [from]. push_back (tmp-2); V [to]. push_back (tmp-1);} bool BFS (int s, int t) {del (VIS,-1); queue <int> q; q. push (s); vis [s] = 0; while (! Q. empty () {int x = Q. front (); q. pop (); For (INT I = 0; I <V [X]. size (); I ++) {node TMP = E [V [x] [I]; If (vis [TMP. to] <0 & TMP. cap> TMP. flow) // The second condition ensures {vis [TMP. to] = vis [x] + 1; q. push (TMP. to) ;}}} if (vis [T]> 0) return true; return false;} int DFS (int o, int F, int T) {If (O = T | f = 0) // returns F; int A = 0, ANS = 0; For (Int & I = cur [O]; I <V [O]. size (); I ++) // note the preceding '&', which is an important optimization {node & TMP = E [V [O] [I]; if (vis [TMP. to] = (vis [O] + 1) & (A = DFS (TMP. to, min (F, TMP. cap-tmp.flow), t)> 0) {TMP. flow + = A; E [V [O] [I] ^ 1]. flow-= A; // save graph mode ans + = A; F-= A; If (F = 0) // note break optimization;} return ans; // optimization} int dinci (int s, int t) {int ans = 0; while (BFS (S, T) {del (cur, 0 ); int TM = DFS (S, INF, T); ans + = TM;} return ans;} void v_clear (int n) {for (INT I = 0; I <= N; I ++) V [I]. clear (); E. clear () ;}int main () {// freopen ("input.txt", "r", stdin); int t; scanf ("% d", & T ); for (int cas = 1; CAS <= T; CAS ++) {int n, m, sum = 0; scanf ("% d", & N, & M); int S = 0, T = N + m + 1, x; For (INT I = 1; I <= N; I ++) scanf ("% d", & X), add_node (S, I, x), sum + = x; For (INT I = 1; I <= m; I ++) scanf ("% d", & X), add_node (n + I, t, x); For (INT I = 1; I <= N; I ++) {scanf ("% d", & X); While (X --) {int TMP; scanf ("% d", & TMP); add_node (I, N + TMP + 1, INF) ;}}for (INT I = 1; I <= m; I ++) {for (Int J = 1; j <= m; j ++) {scanf ("% d", & X); If (x) add_node (n + I, n + J, INF );}} printf ("case # % d: % d \ n", Cas, Sum-dinci (S, T); v_clear (t) ;}return 0 ;}

Hdoj 4971 a simple brute force problem. [maximum closing weight --> Minimum Cut]