【bzoj1066】: [SCOI2007]蜥蜴 图论-最大流

【bzoj1066】: [SCOI2007]蜥蜴

把石柱拆点，流量为高度

然后S与蜥蜴连流量1的边

互相能跳到的石柱连inf的边

石柱能到边界外的和T连inf的边

然后跑dinic就好了

 /* http://www.cnblogs.com/karl07/ */
 #include <cstdlib>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <queue>
 using namespace std;
 const int N=;
 const int inf=1e9;
 struct edge{
     int next,to,c;
 }e[N+];
 int ade=,S=,T=,n,m,c,d,CNT=;
 int first[N+],dis[N+],now[N+],mp[][],I[][],O[][];
 queue <int> Q;

 void addedge(int x,int y,int cap){
     e[++ade].next=first[x];
     e[ade].to=y;
     e[ade].c=cap;
     first[x]=ade;
     e[++ade].next=first[y];
     e[ade].to=x;
     e[ade].c=;
     first[y]=ade;
 }

 #define s e[x].to
 #define cap e[x].c
 #define CAP e[x^1].c
 bool bfs(){
     for (int i=;i<=CNT;i++) dis[i]=inf;
     Q.push(S);
     dis[S]=;
     while (!Q.empty()){
         int p=Q.front();
         Q.pop();
         for (int x=first[p];x;x=e[x].next){
             if (dis[s]>dis[p]+ && cap>){
                 Q.push(s);
                 dis[s]=dis[p]+;
             }
         }
     }
     return dis[T]!=inf;
 }

 int dfs(int p,int mf){
     if (p==T) return mf;
     for (int x=now[p];x;x=e[x].next){
         now[p]=x;
         if (dis[s]==dis[p]+ && cap>){
             int f=dfs(s,min(mf,cap));
             if (f){
                 cap-=f;
                 CAP+=f;
                 return f;
             }
         }
     }
     return ;
 }
 #undef s
 #undef cap
 #undef CAP

 int dinic(){
     int ans=,f;
     while (bfs()){
         for (int i=;i<=CNT;i++) now[i]=first[i];
         while (f=dfs(S,inf)){
             ans+=f;
         }
     }
     return ans;
 }

 void make_p(){
     for (int i=;i<=n;i++){
         for (int j=;j<=m;j++){
             if (mp[i][j]){
                 addedge(I[i][j],O[i][j],mp[i][j]);
                 for (int x=i-d;x<=i+d;x++){
                     for (int y=j-d;y<=j+d;y++){
                         if (abs(i-x)+abs(j-y)>d) continue;
                         if (x< || x>n || y< || y>m) {addedge(O[i][j],T,inf); continue;}
                         if (mp[x][y] && (x!=i || y!=j)){addedge (O[i][j],I[x][y],inf); }
                     }
                 }
             }
         }
     }
 }

 int main(){
     scanf("%d%d%d",&n,&m,&d);
     for (int i=;i<=n;i++){
         char s[];
         scanf("%s",s+);
         for (int j=;j<=m;j++){
             mp[i][j]=s[j]-'';
             if (mp[i][j]) I[i][j]=++CNT,O[i][j]=++CNT;
         }
     }
     make_p();
     for (int i=;i<=n;i++){
         char s[];
         scanf("%s",s+);
         for (int j=;j<=m;j++){
             if (s[j]=='L'){
                 addedge(S,I[i][j],);
                 c++;
             }
         }
     }
     printf("%d\n",c-dinic());
     return ;
 }