Xtu bipartite graph and network flow exercise questions c. Number of squares (1)

C. number of squares (1) Time Limit: 5000 msmemory limit: 32768kb64-bit integer Io format: % i64d Java class name: Main gives you a checkboard with N x n, each grid contains a non-negative number.
Take out several numbers from them so that the lattice of any two numbers does not have a public edge. That is to say, the two grids of the obtained number cannot be adjacent and the sum of the obtained numbers is the largest. Input includes multiple test instances. Each test instance includes an integer N and N x N non-negative (n <= 20). Output for each test instance, maximum possible output and sample input

375 15 21 75 15 28 34 70 5 

Sample output

188

Problem-solving: Some people use status compression DP. I can only knock on others' code .....

1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <cmath> 5 # include <algorithm> 6 # include <climits> 7 # include <vector> 8 # include <queue> 9 # include <cstdlib> 10 # include <string> 11 # include <set> 12 # define ll long long13 # define INF 0x3f3f3f14 using namespace STD; 15 const int maxn = 510; 16 int MP [maxn * maxn], TOT, N, SRC, sink; 17 int C [maxn] [maxn]; 18 bool vis [maxn]; 19 int DF S (int u, int low) {20 if (u = sink) return low; 21 if (vis [u]) return 0; 22 vis [u] = true; 23 For (int v = 0, flow; v <= sink; V ++) {24 if (C [u] [v] & (flow = DFS (V, min (low, C [u] [v]) {25 C [u] [v]-= flow; 26 C [v] [u] + = flow; 27 return flow; 28} 29} 30 return 0; 31} 32 int maxflow () {33 int ans = 0, flow; 34 memset (VIS, false, sizeof (VIS); 35 while (flow = DFS (SRC, INF) {36 memset (VIS, false, sizeof (VIS); 37 ans + = flow; 38} 39 return ans; 40} 41 int main () {42 int I, j, temp; 43 while (~ Scanf ("% d", & N) {44 J = N * n; 45 src = tot = 0; 46 sink = J + 1; 47 memset (MP, 0, sizeof (MP); 48 memset (C, 0, sizeof (c); 49 for (I = 1; I <= J; I ++) {50 scanf ("% d", & temp); 51 tot + = temp; 52 if (I <= N) MP [I] =! MP [I-1]; 53 else MP [I] =! MP [I-n]; 54 if (MP [I]) {55 if (I % N) C [I] [I + 1] = inf; // 56 if (I % n! = 1) C [I] [I-1] = inf; // left 57 if (I <= N * (n-1) c [I] [n + I] = inf; // 58 if (I> N) C [I] [I-n] = inf; // 59 C [SRC] [I] = temp; 60} else C [I] [sink] = temp; 61} 62 printf ("% d \ n", tot-maxflow (); 63} 64 return 0; 65}

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