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。。
1 #include<cstdio> 2 #include<cstring> 3 #include<queue> 4 #include<algorithm> 5 using namespace std; 6 #define INF (1<<30) 7 #define MAXN 888 8 #define MAXM 888*888*2 9 10 struct Edge{ 11 int v,cap,flow,next; 12 }edge[MAXM]; 13 int vs,vt,NE,NV; 14 int head[MAXN]; 15 16 void addEdge(int u,int v,int cap){ 17 edge[NE].v=v; edge[NE].cap=cap; edge[NE].flow=0; 18 edge[NE].next=head[u]; head[u]=NE++; 19 edge[NE].v=u; edge[NE].cap=0; edge[NE].flow=0; 20 edge[NE].next=head[v]; head[v]=NE++; 21 } 22 23 int level[MAXN]; 24 int gap[MAXN]; 25 void bfs(){ 26 memset(level,-1,sizeof(level)); 27 memset(gap,0,sizeof(gap)); 28 level[vt]=0; 29 gap[level[vt]]++; 30 queue<int> que; 31 que.push(vt); 32 while(!que.empty()){ 33 int u=que.front(); que.pop(); 34 for(int i=head[u]; i!=-1; i=edge[i].next){ 35 int v=edge[i].v; 36 if(level[v]!=-1) continue; 37 level[v]=level[u]+1; 38 gap[level[v]]++; 39 que.push(v); 40 } 41 } 42 } 43 44 int pre[MAXN]; 45 int cur[MAXN]; 46 int ISAP(){ 47 bfs(); 48 memset(pre,-1,sizeof(pre)); 49 memcpy(cur,head,sizeof(head)); 50 int u=pre[vs]=vs,flow=0,aug=INF; 51 gap[0]=NV; 52 while(level[vs]<NV){ 53 bool flag=false; 54 for(int &i=cur[u]; i!=-1; i=edge[i].next){ 55 int v=edge[i].v; 56 if(edge[i].cap!=edge[i].flow && level[u]==level[v]+1){ 57 flag=true; 58 pre[v]=u; 59 u=v; 60 //aug=(aug==-1?edge[i].cap:min(aug,edge[i].cap)); 61 aug=min(aug,edge[i].cap-edge[i].flow); 62 if(v==vt){ 63 flow+=aug; 64 for(u=pre[v]; v!=vs; v=u,u=pre[u]){ 65 edge[cur[u]].flow+=aug; 66 edge[cur[u]^1].flow-=aug; 67 } 68 //aug=-1; 69 aug=INF; 70 } 71 break; 72 } 73 } 74 if(flag) continue; 75 int minlevel=NV; 76 for(int i=head[u]; i!=-1; i=edge[i].next){ 77 int v=edge[i].v; 78 if(edge[i].cap!=edge[i].flow && level[v]<minlevel){ 79 minlevel=level[v]; 80 cur[u]=i; 81 } 82 } 83 if(--gap[level[u]]==0) break; 84 level[u]=minlevel+1; 85 gap[level[u]]++; 86 u=pre[u]; 87 } 88 return flow; 89 } 90 char map1[22][22],map2[22][22]; 91 int main(){ 92 int t,n,d; 93 scanf("%d",&t); 94 for(int cse=1; cse<=t; ++cse){ 95 scanf("%d%d",&n,&d); 96 for(int i=0; i<n; ++i) scanf("%s",map1[i]); 97 for(int i=0; i<n; ++i) scanf("%s",map2[i]); 98 99 int tot=0,m=strlen(map1[0]); 100 vs=n*m*2; vt=vs+1; NV=vt+1; NE=0; 101 memset(head,-1,sizeof(head)); 102 103 for(int i=0; i<n; ++i){ 104 for(int j=0; j<m; ++j){ 105 addEdge(i*m+j,i*m+j+n*m,map1[i][j]-'0'); 106 for(int ni=0; ni<n; ++ni){ 107 for(int nj=0; nj<m; ++nj){ 108 if(ni==i && nj==j) continue; 109 if((nj-j)*(nj-j)+(ni-i)*(ni-i)<=d*d) addEdge(i*m+j+n*m,ni*m+nj,INF); 110 } 111 } 112 if(map2[i][j]=='L') addEdge(vs,i*m+j,1),++tot; 113 if(i+1<=d||n-i<=d || j+1<=d||m-j<=d) addEdge(i*m+j+n*m,vt,INF); 114 } 115 } 116 tot-=ISAP(); 117 if(tot==0) printf("Case #%d: no lizard was left behind.\n",cse); 118 else if(tot==1) printf("Case #%d: 1 lizard was left behind.\n",cse); 119 else printf("Case #%d: %d lizards were left behind.\n",cse,tot); 120 } 121 return 0; 122 }
