Number of HDU 1569 squares (2)

Number of squares (2) time limit: 5000 msmemory limit: 32768 kbthis problem will be judged on HDU. Original ID: 1569
64-bit integer Io format: % i64d Java class name: Main gives you a checkboard of M * n, each of which has 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 2 integer m, n, and M * n non-negative numbers (M <= 50, n <= 50). Output for each test instance, maximum possible output and sample input

3 375 15 21 75 15 28 34 70 5

Sample output

188

Sourcehappy 2007 problem solving: Maximum vertex weight independence set = sum-minimum vertex weight independence set = Minimum Cut = maximum stream. Black/white board. I + J is an even point, which is directly connected to a super sink point. The weight is the weight of the Change Point. I + J is an odd vertex connected to a super Source Vertex and its weight is the vertex's weight. In addition, the weights are infinitely large when connected to neighboring four vertices.

 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 #include <stack>13 #define LL long long14 #define pii pair<int,int>15 #define INF 0x3f3f3f3f16 using namespace std;17 const int maxn = 3000;18 struct arc {19     int to,flow,next;20     arc(int x = 0,int y = 0,int z = 0) {21         to = x;22         flow = y;23         next = z;24     }25 };26 arc e[maxn*10];27 int head[maxn],d[maxn],tot;28 int S,T,q[maxn],hd,tail,n,m;29 void add(int u,int v,int flow) {30     e[tot] = arc(v,flow,head[u]);31     head[u] = tot++;32     e[tot] = arc(u,0,head[v]);33     head[v] = tot++;34 }35 bool bfs() {36     for(int i = 1; i <= T; i++) d[i] = 0;37     d[S] = 1;38     hd = tail = 0;39     q[tail++] = S;40     while(hd < tail) {41         int u = q[hd++];42         for(int i = head[u]; ~i; i = e[i].next) {43             if(e[i].flow && !d[e[i].to]) {44                 d[e[i].to] = d[u]+1;45                 q[tail++] = e[i].to;46             }47         }48     }49     return d[T];50 }51 int dfs(int u,int low) {52     int tmp = 0,f;53     if(u == T || !low) return low;54     for(int i = head[u]; ~i; i = e[i].next) {55         if(d[e[i].to] == d[u]+1 && (f = dfs(e[i].to,min(low,e[i].flow)))) {56             tmp += f;57             e[i].flow -= f;58             e[i^1].flow += f;59             low -= f;60             if(!low) break;61         }62     }63     return tmp;64 }65 int main() {66     while(~scanf("%d %d",&n,&m)) {67         memset(head,-1,sizeof(head));68         int sum = tot = 0,o,ans = 0;69         S = n*m+1;70         T = n*m+2;71         for(int i = 1; i <= n; i++)72             for(int j = 1; j <= m; j++) {73                 scanf("%d",&o);74                 sum += o;75                 int tmp = (i-1)*m + j;76                 if((i+j)&1) {77                     add(S,tmp,o);78                     if(i > 1) add(tmp,tmp-m,INF);79                     if(j > 1) add(tmp,tmp-1,INF);80                     if(i < n) add(tmp,tmp+m,INF);81                     if(j < m) add(tmp,tmp+1,INF);82                 } else add(tmp,T,o);83             }84         while(bfs()) ans += dfs(S,INF);85         printf("%d\n",sum-ans);86     }87     return 0;88 }

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Number of HDU 1569 squares (2)