Hdu 3657 least cut active/odd/Even square fetch typical questions/least cut

Hdu 3657 least cut active/odd/Even square fetch typical questions/least cut

Question: Take the number of squares. If the number is adjacent, pay a certain price. (Cost: 2 * (X & Y) (start to use the fee stream, knock the upgrade version 3820, kneel down ...)

Graph creation: for adjacent problems, the classic method is to create a bipartite graph for parity. For the adjacent two-point edge 2 * (X & Y), source-> X edge, Y-> sink edge, the weight w is the vertex weight.

Ans = total Point Weight-minimum cut: if the cut edge is source-> X, it indicates that x is not selected (it is cut edge, and the value must be the smallest in the path ), if the cut edge is a Y-sink point, the same is true. If the cut edge is X-> Y, the Y point is selected and the X point is selected, split to w (2 * (X & Y )).

I do not understand its essence. I don't know why. (I started to knock a few more else and kept kneeling !)

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       using namespace std;const int inf=0x3f3f3f3f;const int maxv=2550,maxe=20000;int nume=0;int head[maxv];int e[maxe][3];void inline adde(int i,int j,int c){ e[nume][0]=j;e[nume][1]=head[i];head[i]=nume; e[nume++][2]=c; e[nume][0]=i;e[nume][1]=head[j];head[j]=nume; e[nume++][2]=0;}int ss,tt,n,m,k;int vis[maxv];int lev[maxv];bool bfs(){ for(int i=0;i
       
        q; q.push(ss); vis[ss]=1; while(!q.empty()) { int cur=q.front(); q.pop(); for(int i=head[cur];i!=-1;i=e[i][1]) { int v=e[i][0]; if(!vis[v]&&e[i][2]>0) { lev[v]=lev[cur]+1; vis[v]=1; q.push(v); } } } return vis[tt];}int dfs(int u,int minf){ if(u==tt||minf==0)return minf; int sumf=0,f; for(int i=head[u];i!=-1&&minf;i=e[i][1]) { int v=e[i][0]; if(lev[v]==lev[u]+1&&e[i][2]>0) { f=dfs(v,minf
        
         =0) adde(i*m+j,(i-1)*m+j,2*(mapp[i][j]&mapp[i-1][j])); if(i+1
         
          =0) adde(i*m+j,i*m+j-1,2*(mapp[i][j]&mapp[i][j-1])); if(j+1
          
           %d:%d\n",i,e[j][0],e[j][2]); }*/}void init(){ nume=0;sums=0; ss=n*m+2;tt=ss+1; for(int i=0;i<=tt;i++) { head[i]=-1; must[i]=0; }}int main(){ while(scanf("%d%d%d",&n,&m,&k)!=EOF) { init(); read_build(); int ans; ans=sums-dinic(); printf("%d\n",ans); } return 0;}