POJ2711 Leapin' Lizards（最大流）

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 ;
 }