目录
BZOJ1066 蜥蜴
题解
比较裸的一道网络流，我们先进行拆点，对于每一根柱子，从这根柱子的入点向出点连一条流量为这条柱子的高度的边，对于两根能够相互到达的柱子，连一条流量为\(inf\)的边，然后对于初始有蜥蜴的柱子，从源点向这根柱子连一条流量为1的边，对于所有能够跳出边界的柱子，连一条流量为\(inf\)的边。最后跑一边最大流即可。
ｃｏｄｅ
#includeusing namespace std;typedef long long ll;bool Finish_read;template inline void read(T &x){Finish_read=0;x=0;int f=1;char ch=getchar();while(!isdigit(ch)){if(ch=='-')f=-1;if(ch==EOF)return;ch=getchar();}while(isdigit(ch))x=x*10+ch-'0',ch=getchar();x*=f;Finish_read=1;}template inline void print(T x){if(x/10!=0)print(x/10);putchar(x%10+'0');}template inline void writeln(T x){if(x<0)putchar('-');x=abs(x);print(x);putchar('\n');}template inline void write(T x){if(x<0)putchar('-');x=abs(x);print(x);}/*================Header Template==============*/#define PAUSE printf("Press Enter key to continue..."); fgetc(stdin);const int maxn=50;const int N=1e4+500;const int M=1e5+500;const int inf=0x3f3f3f3f;int n,m,d,cnt,S=0,T=801;int mp[maxn][maxn],idx[maxn][maxn],lz[maxn][maxn];struct edge { int to,nxt,w;}E[M];int tot=1;char s[maxn];int head[N],dis[N];/*==================Define Area================*/void addedge(int u,int v,int w) { E[++tot].to=v;E[tot].nxt=head[u];head[u]=tot;E[tot].w=w; E[++tot].to=u;E[tot].nxt=head[v];head[v]=tot;E[tot].w=0;}int Dis(int x,int y,int nx,int ny) { return abs(x-nx)+abs(y-ny);}bool Judge(int x,int y,int nx,int ny) { if(Dis(x,y,nx,ny)<=d&&mp[x][y]&&mp[nx][ny]) return 1; return 0;}void Build() { for(int i=1;i<=n;i++){ for(int j=1;j<=m;j++) { for(int nx=i-d;nx<=i+d;nx++) { for(int ny=j-d;ny<=j+d;ny++) { if(Judge(i,j,nx,ny)&&(i!=nx||j!=ny)) addedge(idx[i][j]+400,idx[nx][ny],inf); } } } } for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { if(mp[i][j]) addedge(idx[i][j],idx[i][j]+400,mp[i][j]); } } for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { if(lz[i][j]) addedge(S,idx[i][j],1); } } for(int i=1;i<=d;i++) { for(int j=d+1;j<=n-d;j++) { addedge(idx[j][i]+400,T,inf); addedge(idx[j][m-i+1]+400,T,inf); } } for(int i=1;i<=d;i++) { for(int j=1;j<=m;j++) { addedge(idx[i][j]+400,T,inf); addedge(idx[n-i+1][j]+400,T,inf); } }}bool bfs() { queue Q; memset(dis,-1,sizeof dis); Q.push(S); dis[S]=0; while(!Q.empty()) { int o=Q.front(); Q.pop(); for(int i=head[o];~i;i=E[i].nxt) { int to=E[i].to; if(dis[to]==-1&&E[i].w) { dis[to]=dis[o]+1; Q.push(to); } } } return dis[T]!=-1;}int dfs(int o,int flow) { if(o==T) return flow; int used=0,k; for(int i=head[o];~i;i=E[i].nxt) { int to=E[i].to; if(dis[to]==dis[o]+1&&E[i].w) { k=dfs(to,min(flow-used,E[i].w)); E[i].w-=k;E[i^1].w+=k; used+=k; if(used==flow) return flow; } } if(!used) dis[o]=-1; return used;}int main() { memset(head,-1,sizeof head); read(n);read(m);read(d); cnt=0; for(int i=1;i<=n;i++) { scanf("%s",s+1); for(int j=1;j<=m;j++) { mp[i][j]=s[j]-'0'; idx[i][j]=++cnt; } } int c=0; for(int i=1;i<=n;i++) { scanf("%s",s+1); for(int j=1;j<=m;j++) { lz[i][j]=(s[j]=='L'); c+=lz[i][j]; } } Build(); int ans=c; while(bfs()) { ans-=dfs(S,0x3f3f3f3f); } printf("%d\n",ans); return 0;}/*5 8 20000000002000000003211000200000000000000..................LLLL..................*/