BZOJ2132 Rodeo Plan

BZOJ2132 Rodeo Plan

Test instructions

A piece of land can be divided into small areas of nxm blocks. Expire I row J column area, the construction business district will receive the AIJ income, the construction industrial area will obtain the bij income. If the region (I,J) is adjacent (adjacent to two squares with common edges) with a K block (obviously K not more than 4) the type is different from (i,j) area, then this area can increase KXCIJ revenue. The known gain matrix A,b,c, seeking the maximum value of income.

Exercises

Because the additional benefit is not between 22, consider dividing the problem by 2. The entire board is dyed in black and white due to the fact that two blocks of land are required for different properties. If a piece of land A is black, then s->a:c[a] a->t:c[a workers], if White is the opposite s->a:c[a workers] a->t:c[a), for the adjacent ab A<->b c[a and]+c[b together]. The final answer is all the gains and the least cuts.

Code:

1#include <cstdio>2#include <cstring>3#include <algorithm>4#include <queue>5 #defineInc (I,J,K) for (int i=j;i<=k;i++)6 #defineVisit (I,J) for (int i=g[j];i!=-1;i=es[i].n)7 #defineINF 0X3FFFFFFF8 using namespacestd;9 Ten structe{intT,c,n;}; E es[500000];intess,g[20000]; OneInlinevoidPeintFintTintc) {es[++ess]= (e) {t,c,g[f]}; g[f]=ess; es[++ess]= (e) {f,0, G[t]}; g[t]=ess;} AInlinevoidPe2 (intFintTintc) {es[++ess]= (e) {t,c,g[f]}; g[f]=ess; es[++ess]= (e) {f,c,g[t]}; g[t]=ess;} -InlinevoidInit () {ess=-1; memset (g,-1,sizeof(g));} -Queue <int> q;inth[20000]; the BOOLBFsintSintt) { -      while(! Q.empty ()) Q.pop (); memset (h,-1,sizeof(h)); Q.push (s); h[s]=0; -      while(!Q.empty ()) { -         intx=Q.front (); Q.pop (); +Visit (i,x)if(es[i].c&&h[es[i].t]==-1) {h[es[i].t]=h[x]+1; Q.push (es[i].t);} -     } +     if(h[t]==-1)return 0;Else return 1; A } at intDfsintXintTintf) { -     if(x==t)returnFintu=0; -Visit (i,x)if(es[i].c&&h[es[i].t]==h[x]+1){ -         intw=Dfs (Es[i].t,t,min (F,ES[I].C)); -F-=w; U+=w; Es[i].c-=w; es[i^1].c+=w;if(f==0)returnu; -     } in     if(!u) h[x]=-1;returnu; - } to intDinic (intSintT) {intf=0; while(BFS (s,t)) F+=dfs (S,t,inf);returnF;} + inta1[ $][ $],a2[ $][ $],a3[ $][ $],tot,n,m,s,t;BOOLcol[ $][ $]; -InlineintCgintXintY) {return(X-1) *m+y;} the intMain () { *scanf"%d%d", &n,&m); tot=0; $Inc (I,1, N) Inc (J,1, m) scanf ("%d", &a1[i][j]), tot+=A1[i][j];Panax NotoginsengInc (I,1, N) Inc (J,1, m) scanf ("%d", &a2[i][j]), tot+=A2[i][j]; -Inc (I,1, N) Inc (J,1, m) scanf ("%d",&a3[i][j]); thes=0; t=n*m+1; Init (); Inc (I,1, N) Inc (J,1, m) {col[i][j]= (i+j) &1;} +Inc (I,1, N) Inc (J,1, M) { A         if(Col[i][j]) PE (S,CG (I,J), A1[i][j]), PE (CG (I,J), t,a2[i][j]);Elsepe (S,CG (I,J), A2[i][j]), PE (CG (I,J), t,a1[i][j]); the         if(i!=n) Pe2 (CG (I,J), CG (i+1, j), a3[i][j]+a3[i+1][J]), tot+= (a3[i][j]+a3[i+1][j]); +         if(j!=m) Pe2 (CG (I,J), CG (i,j+1), a3[i][j]+a3[i][j+1]), tot+= (a3[i][j]+a3[i][j+1]); -     } $printf"%d", tot-dinic (s,t)); $     return 0; -}

20160324

BZOJ2132 Rodeo Plan