Bzoj 2127 Happiness "Network Flow Dinic"

Reference: Https://www.cnblogs.com/chenyushuo/p/5144957.html has to say that this is a very good way to build a map.
Assuming that S-point selection, T-point selection, a[i][j] for (i,j) election proceeds, b[i][ J] for (I,J) election proceeds, C[i][j] for the simultaneous selection of proceeds, D[i][j] for the simultaneous selection of the proceeds.
For each point x= (i+1) *m+j, we connect
\[c[s,x]=b[i][j] \]
\[c[x,t]=a[i][j] \]
for interest-related X, y two points, connect
\[C[S,X]=D[I][J]/2 \]
\ [C[S,Y]=D[I][J]/2 \]
\[C[X,T]=C[I][J]/2 \]
\[C[Y,T]=C[I][J]/2 \]
\[C[X,Y]=C[I][J]/2+D[I][J]/2 \]
\[c[ Y,X]=C[I][J]/2+D[I][J]/2 \]
Completed diagram:

and then consider the minimum cut, the following enumeration of several cases:
is selected, that is, cut the X-selection, Y-selected Rie (x and y) are selected:

are selected, Cut out the X-selection, y selected text and (x, y) are selected text:

x selection y, that is, cut the x selection, y selection, (x, y) are selected/+ (x, y) are selected/2+ (x, y) are selected/2+ (x, y) is the choice of the text/2, that is, cut the x selection, y selection, All options, (x, y) are selected:

y selection x, that is, cut off the x selection, y selection, (x, y) are selected/+ (x, y) are selected/2+ (x, y) are selected/2+ (x, y) is the choice of the text/2, that is, cut the x selection, y selection, (x, y) are selected, (x, Y Tu Xuanwen:

For the operation divided by 2, in order to avoid the error of the next rounding, we choose to take all the traffic, and finally/2.
$ ans=sum (full benefit)-min cut $
P.S. When using the adjacency table to build a map, the first point of the menu section of the edge connected, and then to practice the same time selected revenue, otherwise it will be heavy

#include <iostream>#include <cstdio>#include <cstring>#include <queue>using namespaceStdConst intn= the, e=1000005, inf=1e9, p=500005;intN,m,a[n][n],b[n][n],c[n][n],d[n][n],s,t,sum,cnt=1, H[p],le[n*n];structqwe{intNe,to,va;} E[e];intRead () {intR=0;CharP=getchar (); while(p<' 0 '|| P>' 9 ') P=getchar (); while(p>=' 0 '&&p<=' 9 ') {r=r*Ten+p-48;    P=getchar (); }returnt;;voidAddintUintVintW) {cnt++;    E[cnt].ne=h[u];    E[cnt].to=v;    E[cnt].va=w; h[u]=cnt;}BOOLBFS () {queue<int>q; Memset (LE,0,sizeof(LE)); le[s]=1; Q.push (s); while(!q.empty ()) {intU=q.front ();//cout<<u<< "Aaaaaaaaaaaa" <<endl;Q.pop (); for(intI=h[u];i;i=e[i].ne)if(E[i].va&&!le[e[i].to]) {Le[e[i].to]=le[u]+1;            Q.push (e[i].to); }    }//for (int i=0;i<=5;i++)        //cout<<le[i]<< "";    returnLE[T]; }intDfsintUintf) {//cout<<u<< "" <<f<<endl;    if(u==t) {//cout<< "!";        returnF }intUsed=0; for(intI=h[u];i;i=e[i].ne) {//cout<<u<< "" <<e[i].to<< "" <<e[i].va<<endl;;        if(e[i].va>0&&le[e[i].to]==le[u]+1)        {//cout<< "OK" <<endl;            intW=dfs (E[i].to,min (F-used,e[i].va));            E[i].va-=w; e[i^1].va+=w; Used+=w;if(used==f)returnF }    }if(!used) le[u]=-1;returnUsed;}intDinic () {intans=0; while(BFS ())) Ans+=dfs (S,inf);//,cout<<ans<<endl;    returnAns;}intMain () {N=read (), M=read (); s=0, t=n*m+1; for(intI=1; i<=n;i++) for(intj=1; j<=m;j++) A[i][j]=read (), sum+=a[i][j]; for(intI=1; i<=n;i++) for(intj=1; j<=m;j++) B[i][j]=read (), sum+=b[i][j]; for(intI=1; i<=n;i++) for(intj=1; j<=m;j++) {intx= (i-1) *m+j; Add (s,x,b[i][j]<<1); Add (X,s,0); Add (x,t,a[i][j]<<1); Add (T,x,0); } for(intI=1; i<n;i++) for(intj=1; j<=m;j++) C[i][j]=read (), sum+=c[i][j]; for(intI=1; i<n;i++) for(intj=1; j<=m;j++) D[i][j]=read (), sum+=d[i][j]; for(intI=1; i<n;i++) for(intj=1; j<=m;j++) {intx= (i-1) *m+j,y=i*m+j;            Add (S,x,d[i][j]); Add (X,s,0);            Add (S,y,d[i][j]); Add (Y,s,0);            Add (X,y,c[i][j]+d[i][j]);            Add (Y,x,c[i][j]+d[i][j]);            Add (X,t,c[i][j]); Add (T,x,0);            Add (Y,t,c[i][j]); Add (T,y,0); } for(intI=1; i<=n;i++) for(intj=1; j<m;j++) C[i][j]=read (), sum+=c[i][j]; for(intI=1; i<=n;i++) for(intj=1; j<m;j++) D[i][j]=read (), sum+=d[i][j]; for(intI=1; i<=n;i++) for(intj=1; j<m;j++) {intx= (i-1) *m+j,y= (i-1) *m+j+1;            Add (S,x,d[i][j]); Add (X,s,0);            Add (S,y,d[i][j]); Add (Y,s,0);            Add (X,y,c[i][j]+d[i][j]);            Add (Y,x,c[i][j]+d[i][j]);            Add (X,t,c[i][j]); Add (T,x,0);            Add (Y,t,c[i][j]); Add (T,y,0); } printf ("%d\n", sum-(Dinic () >>1));return 0;}

Bzoj 2127 Happiness "Network Flow Dinic"