hdu2732最大流

這道題目就是拆點，給出是每個點的容量限制，每個點如果可以escape就對匯點連條inf，每個點對可到的柱子連條inf，源點到有lizard的地方連條長為1的邊，ans就是sum（lizard)-最大流

#include<cstdio>#include<cstring>int T,n,d,m,cnt,dis[1000],gap[1000],head[1000],s,t,sum,N;const int inf=10000000;struct EDGE{int to,f,nxt;}edge[300000];char g[30][30],gg[30][30],x[30][30];int dx[]={1,1,-1,-1};int dy[]={1,-1,1,-1};int min(int a,int b){return a<b?a:b;}void add(int x,int y,int c){edge[cnt].to=y;edge[cnt].f=c;edge[cnt].nxt=head[x];head[x]=cnt++;}void gao(int x,int y,int c){add(x,y,c);add(y,x,0);}int safe(int a,int b){return a-d<=0||a+d>=n+1||b-d<=0||b+d>=m+1;}int in(int a,int b){return a>=1&&a<=n&&b>=1&&b<=m;}void init(){memset(head,-1,sizeof(head));memset(dis,0,sizeof(dis));memset(gap,0,sizeof(gap));sum=cnt=0;}int dfs(int x,int flow){if(x==t)return flow;int temp=flow;int pos=t-1;int j;for(j=head[x];j!=-1;j=edge[j].nxt){int y=edge[j].to;int c=edge[j].f;if(c>0&&dis[x]==dis[y]+1){int temp_flow=dfs(y,min(temp,c));temp-=temp_flow;edge[j].f-=temp_flow;edge[j^1].f+=temp_flow;if(temp==0||dis[s]==t)return flow-temp;}if(c>0&&pos>dis[y])pos=dis[y];}if(temp==flow){if(!(--gap[dis[x]]))dis[s]=t;elsegap[dis[x]=pos+1]++;}return flow-temp;}int sap(){int maxflow=0;gap[0]=t;while(dis[s]<t){maxflow+=dfs(s,inf);}return sum-maxflow;}int main(){int pro=0;scanf("%d",&T);while(T--){scanf("%d%d",&n,&d);init();for(int i=1;i<=n;i++)scanf("%s",g[i]+1);m=strlen(g[1]+1);N=n*m;s=2*N+1,t=s+1;for(int i=1;i<=n;i++)scanf("%s",gg[i]+1);int k1=0,k2=0;for(int i=1;i<=n;i++)for(int j=1;j<=m;j++){int num=(i-1)*m+j;if(safe(i,j))gao(num+N,t,inf);if(g[i][j]-'0')gao(num,num+N,g[i][j]-'0');}for(int i=1;i<=n;i++)for(int j=1;j<=m;j++){if(gg[i][j]=='L'){int num=(i-1)*m+j;sum++;gao(s,num,1);}}for(int i=1;i<=n;i++)for(int j=1;j<=m;j++){int num=(i-1)*m+j;for(int z=0;z<4;z++)for(int k1=0;k1<=d;k1++)for(int k2=0;k2<=d;k2++)if(k1+k2>0&&k1+k2<=d){int x=i+k1*dx[z];int y=j+k2*dy[z];int mun=(x-1)*m+y;if(in(x,y))gao(num+N,mun,inf);}}int ans=sap();if(ans>1)printf("Case #%d: %d lizards were left behind.\n",++pro,ans);          else if(ans)printf("Case #%d: %d lizard was left behind.\n",++pro,ans);          else printf("Case #%d: no lizard was left behind.\n",++pro);  }return 0;}