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很普通的拆点网络流，把每个柱子拆成两个点(i,j,0)和(i,j,1).对于柱子的高度限制则加边((i,j,0),(i,j,1),height).
两个柱子能互相到达则加边((i,j,1),(i1,j1,0),INF). 能到达边界的柱子加边((i,j,1),t,INF).有蜥蜴的柱子加边(s,(i,j,0),1).
跑一遍最大流，答案就是总蜥蜴数-最大流。
# include # include # include # include # include # include # include # include # include # include # include using namespace std;# define lowbit(x) ((x)&(-x))# define pi acos(-1.0)# define eps 1e-9# define MOD 12345678# define INF 1000000000# define mem(a,b) memset(a,b,sizeof(a))# define FOR(i,a,n) for(int i=a; i<=n; ++i)# define FO(i,a,n) for(int i=a; i PII;typedef vector VI;# pragma comment(linker, "/STACK:1024000000,1024000000")typedef long long LL;int Scan() { int res=0, flag=0; char ch; if((ch=getchar())=='-') flag=1; else if(ch>='0'&&ch<='9') res=ch-'0'; while((ch=getchar())>='0'&&ch<='9') res=res*10+(ch-'0'); return flag?-res:res;}void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0');}const int N=2005;//Code begin... struct Edge{ int p, next, w;}edge[650005];int head[805], cnt=2, s, t, vis[805];char s1[25][25], s2[25][25];queue Q; void add_edge(int u, int v, int w){ edge[cnt].p=v; edge[cnt].next=head[u]; edge[cnt].w=w; head[u]=cnt++; edge[cnt].p=u; edge[cnt].next=head[v]; edge[cnt].w=0; head[v]=cnt++;}int bfs(void){ int i, v; mem(vis,-1); while (!Q.empty()) Q.pop(); vis[s]=0; Q.push(s); while (!Q.empty()) { v=Q.front(); Q.pop(); for (i=head[v]; i; i=edge[i].next) { if (edge[i].w>0 && vis[edge[i].p]==-1) { vis[edge[i].p]=vis[v] + 1; Q.push(edge[i].p); } } } return vis[t]!=-1;}int dfs(int x, int low){ int i, a, temp=low; if (x==t) return low; for (i=head[x]; i; i=edge[i].next) { if (edge[i].w>0 && vis[edge[i].p] == vis[x] + 1){ a=dfs(edge[i].p,min(edge[i].w,temp)); temp-=a; edge[i].w-=a; edge[i^1].w += a; if (temp==0) break; } } if (temp==low) vis[x]=-1; return low-temp;}int main (){ int n, m, d, sum=0; scanf("%d%d%d",&n,&m,&d); FO(i,0,n) scanf("%s",s1[i]+1); FO(i,0,n) scanf("%s",s2[i]+1); s=0, t=2*n*m+1; FO(i,0,n) FOR(j,1,m) { if (s1[i][j]!='0') add_edge(i*m+j,i*m+j+n*m,s1[i][j]-'0'); if (s2[i][j]=='L') add_edge(s,i*m+j,1), ++sum; if (i <=d||(m-j+1)<=d) add_edge(i*m+j+n*m,t,INF); FO(k,0,n) FOR(l,1,m) { if (k==i&&l==j) continue; if (abs(k-i)+abs(l-j)<=d) add_edge(i*m+j+n*m,k*m+l,INF); } } int res=0, temp; while (bfs()) while (temp=dfs(s,INF)) res+=temp; printf("%d\n",sum-res); return 0;}
转载于:https://www.cnblogs.com/lishiyao/p/6502431.html
