比较形象的是地图每个点都拆成三个点，这三个点限制流量为0或1，于是再一分为二，这样每个点都被拆成6个点。。。
其实拆两个点，连容量为柱子高的边，这样就行了。。
这题我掉坑了，“1 lizard was left behind.”。。虽然样例都把一切都说了。。要注意细节。。
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
#define INF (1<<30)
#define MAXN 888
#define MAXM 888*888*2 struct Edge{
int v,cap,flow,next;
}edge[MAXM];
int vs,vt,NE,NV;
int head[MAXN]; void addEdge(int u,int v,int cap){
edge[NE].v=v; edge[NE].cap=cap; edge[NE].flow=;
edge[NE].next=head[u]; head[u]=NE++;
edge[NE].v=u; edge[NE].cap=; edge[NE].flow=;
edge[NE].next=head[v]; head[v]=NE++;
} int level[MAXN];
int gap[MAXN];
void bfs(){
memset(level,-,sizeof(level));
memset(gap,,sizeof(gap));
level[vt]=;
gap[level[vt]]++;
queue<int> que;
que.push(vt);
while(!que.empty()){
int u=que.front(); que.pop();
for(int i=head[u]; i!=-; i=edge[i].next){
int v=edge[i].v;
if(level[v]!=-) continue;
level[v]=level[u]+;
gap[level[v]]++;
que.push(v);
}
}
} int pre[MAXN];
int cur[MAXN];
int ISAP(){
bfs();
memset(pre,-,sizeof(pre));
memcpy(cur,head,sizeof(head));
int u=pre[vs]=vs,flow=,aug=INF;
gap[]=NV;
while(level[vs]<NV){
bool flag=false;
for(int &i=cur[u]; i!=-; i=edge[i].next){
int v=edge[i].v;
if(edge[i].cap!=edge[i].flow && level[u]==level[v]+){
flag=true;
pre[v]=u;
u=v;
//aug=(aug==-1?edge[i].cap:min(aug,edge[i].cap));
aug=min(aug,edge[i].cap-edge[i].flow);
if(v==vt){
flow+=aug;
for(u=pre[v]; v!=vs; v=u,u=pre[u]){
edge[cur[u]].flow+=aug;
edge[cur[u]^].flow-=aug;
}
//aug=-1;
aug=INF;
}
break;
}
}
if(flag) continue;
int minlevel=NV;
for(int i=head[u]; i!=-; i=edge[i].next){
int v=edge[i].v;
if(edge[i].cap!=edge[i].flow && level[v]<minlevel){
minlevel=level[v];
cur[u]=i;
}
}
if(--gap[level[u]]==) break;
level[u]=minlevel+;
gap[level[u]]++;
u=pre[u];
}
return flow;
}
char map1[][],map2[][];
int main(){
int t,n,d;
scanf("%d",&t);
for(int cse=; cse<=t; ++cse){
scanf("%d%d",&n,&d);
for(int i=; i<n; ++i) scanf("%s",map1[i]);
for(int i=; i<n; ++i) scanf("%s",map2[i]); int tot=,m=strlen(map1[]);
vs=n*m*; vt=vs+; NV=vt+; NE=;
memset(head,-,sizeof(head)); for(int i=; i<n; ++i){
for(int j=; j<m; ++j){
addEdge(i*m+j,i*m+j+n*m,map1[i][j]-'');
for(int ni=; ni<n; ++ni){
for(int nj=; nj<m; ++nj){
if(ni==i && nj==j) continue;
if((nj-j)*(nj-j)+(ni-i)*(ni-i)<=d*d) addEdge(i*m+j+n*m,ni*m+nj,INF);
}
}
if(map2[i][j]=='L') addEdge(vs,i*m+j,),++tot;
if(i+<=d||n-i<=d || j+<=d||m-j<=d) addEdge(i*m+j+n*m,vt,INF);
}
}
tot-=ISAP();
if(tot==) printf("Case #%d: no lizard was left behind.\n",cse);
else if(tot==) printf("Case #%d: 1 lizard was left behind.\n",cse);
else printf("Case #%d: %d lizards were left behind.\n",cse,tot);
}
return ;
}